Overview
Systems of linear inequalities are one of the more conceptually demanding Algebra topics on the SAT, typically appearing in the harder module. The SAT tests three main skills: identifying which region on a graph satisfies a system of inequalities, determining whether a specific point satisfies the system, and writing or interpreting a system that models real-world constraints. The feasible region — the overlap of all shaded half-planes — is the central concept, and questions frequently test whether students correctly identify the boundary line type (solid vs. dashed) and the direction of shading.
Key Points
1. Graphing a Single Linear Inequality
Before working with systems, master graphing one inequality:
Step 1: Rewrite in slope-intercept form y = mx + b
Step 2: Graph the boundary line
Dashed line ──── for < or > (points on line are NOT solutions)
Solid line ━━━━ for ≤ or ≥ (points on line ARE solutions)
Step 3: Determine which side to shade
y > mx + b or y ≥ mx + b → shade ABOVE the line
y < mx + b or y ≤ mx + b → shade BELOW the line
Step 4: Verify with the test point (0,0) if it is not on the line
2. Graphing a System of Linear Inequalities
A system of linear inequalities is two or more inequalities that must all be satisfied simultaneously.
Process:
Step 1: Graph each inequality separately (draw boundary, shade half-plane)
Step 2: The FEASIBLE REGION is the area where ALL shaded regions overlap
Step 3: Any point in the feasible region satisfies every inequality in the system
Boundary line summary:
| Inequality symbol | Boundary line type | Points on line included? |
|---|---|---|
| < or > | Dashed | No |
| ≤ or ≥ | Solid | Yes |
3. The Test Point Method
The test point method is the most reliable way to determine shading direction and to verify solutions.
Step 1: Convert each inequality to slope-intercept form
Step 2: Graph boundary lines
Step 3: Choose a test point NOT on any boundary line
Best choice: (0, 0) — easiest arithmetic, as long as no boundary line passes through origin
Step 4: Substitute the test point into each inequality separately
Step 5: If the test point satisfies the inequality → shade the side containing (0, 0)
If the test point does NOT satisfy → shade the OPPOSITE side
Step 6: The feasible region is where ALL shaded regions overlap
Example:
System: y < -x + 4 and y ≥ 2x - 1
Test (0, 0):
0 < -0 + 4 → 0 < 4 ✓ shade the side containing (0,0) for first inequality
0 ≥ 2(0) - 1 → 0 ≥ -1 ✓ shade the side containing (0,0) for second inequality
(0, 0) is in the feasible region.
4. Testing Whether a Point Satisfies the System
SAT multiple-choice questions often give four specific points and ask which one satisfies the system. The most efficient approach:
For each answer choice (x, y):
Step 1: Substitute into inequality 1 — is it true?
Step 2: Substitute into inequality 2 — is it true?
Step 3: The correct answer satisfies ALL inequalities simultaneously
This is almost always faster than graphing on the multiple-choice format.
Key insight from SAT problems: If (0, 0) is given as a solution to y < -x + a and y > x + b:
Substitute (0,0): 0 < a → a > 0
0 > b → b < 0
This type of algebraic reasoning about signs of constants is a common SAT higher-difficulty question.
5. Real-World Constraint Problems
The SAT presents systems of inequalities as multi-constraint optimization or feasibility problems.
Framework:
Step 1: Identify each constraint in the word problem
Step 2: Write one inequality per constraint
(translate "at most", "no more than" → ≤; "at least" → ≥)
Step 3: Graph the system or test specific answer choices
Step 4: The feasible region contains all (x, y) values that satisfy all constraints
Example:
"A store sells apples (x) at $2 each and oranges (y) at $3 each.
The store can hold at most 100 pieces of fruit.
The store wants to earn at least $200."
Constraints:
x + y ≤ 100 (capacity)
2x + 3y ≥ 200 (revenue)
x ≥ 0, y ≥ 0 (non-negative)
6. Desmos on the Digital SAT
Desmos handles systems of inequalities natively — enter each inequality and the feasible region (overlap) is shaded automatically. This is the fastest method for identifying feasible regions when exact boundary values are given.
Pitfalls and Common Mistakes
Mistake 1: Shading the wrong side of the boundary line
- Wrong approach: For y > 2x + 1, shade below the line
- Correct approach: y > 2x + 1 means y is GREATER THAN the right side → shade ABOVE the line
- Fix: After converting to slope-intercept form, the inequality direction (< vs >) directly tells you above or below. Confirm with test point (0,0).
Mistake 2: Using a dashed line when the inequality includes equality
- Wrong approach: Drawing a dashed line for y ≤ 3x - 2
- Correct approach: ≤ includes the boundary → draw a SOLID line
- Fix: Check the inequality symbol: strict (< >) = dashed; non-strict (≤ ≥) = solid. This affects whether points on the boundary count as solutions.
Mistake 3: Only checking one inequality when testing a point
- Wrong approach: Point (2, 3) satisfies y > x + 1, so it’s the answer (without checking the second inequality)
- Correct approach: Check (2, 3) in EVERY inequality in the system — it must satisfy all of them
- Fix: Go through each inequality one by one for every candidate point. If any inequality fails, eliminate that point.
Mistake 4: Not converting to slope-intercept form before shading
- Wrong approach: 3x - 2y > 6, shade where x > 2 (incorrectly treating coefficients)
- Correct approach: Rewrite: -2y > -3x + 6 → y < (3/2)x - 3 (flip sign when dividing by -2) → shade BELOW
- Fix: Always isolate y first. Remember to flip the inequality if dividing by a negative coefficient.
Related Entries
- Linear_Inequalities — Single-variable inequalities (foundation)
- Systems_Linear_Equations — Systems without inequality — comparison of concepts
- Linear_Functions_Graphs — Graphing boundary lines
- Absolute_Value — Absolute value inequalities also create two-region solutions
- Linear_Equations — Equation-solving foundational skills
Quick Reference Card
| Concept | Rule |
|---|---|
| Boundary line — dashed | < or > (points NOT included) |
| Boundary line — solid | ≤ or ≥ (points ARE included) |
| Shade above | y > mx + b or y ≥ mx + b |
| Shade below | y < mx + b or y ≤ mx + b |
| Feasible region | Overlap of ALL shaded half-planes |
| Test point method | Plug (0,0) into each inequality; shade accordingly |
| ”at least” | ≥ |
| “at most” | ≤ |
| Testing a specific point | Substitute into ALL inequalities — must satisfy every one |
| Desmos | Enter inequalities directly; overlap region appears automatically |
| Convert sign flip | Dividing by negative when isolating y → flip inequality |