How to Solve Systems of Linear Inequalities?
Systems of Linear Inequalities
A system of linear inequalities is just two (or more) inequalities graphed together — and the solution is the region where their shaded areas overlap. Graph each boundary, shade each side, and the double-shaded zone is your answer. We’ll do it with a test point, plus a solver, drills, and a worksheet maker a tap away.
A system of linear inequalities takes the “shade a region” idea from a single inequality and stacks two of them together. Each inequality shades part of the plane; the solution to the system is the area where those shaded parts overlap. Graph each one, then look for the double-shaded zone — every point there satisfies both at once.
In short: graph each inequality (dashed or solid boundary, shade the correct side), and the solution is the region where the shadings overlap. A point is a solution only if it makes both inequalities true.
Two Regions, One Overlap
Each linear inequality splits the plane into a “true” half and a “false” half. With two inequalities you get two true-halves; the system’s solution is their intersection — the slice of the plane shaded by both. A quick test point confirms whether a spot is in that overlap.
How to solve (3 steps):
- Graph each boundary line — dashed for \(<\)/\(>\), solid for \(\le\)/\(\ge\).
- Shade the correct side of each (use a test point).
- The overlap of the two shadings is the solution region.
\(y < 2x + 1\) (and \(y \ge -x + 2\))
Below is the region for \(y < 2x + 1\) — a dashed boundary with the side away from the origin shaded. The full system's answer is where this overlaps the region for \(y \ge -x + 2\). The point \((3,0)\), for instance, satisfies both.
⚡ Graph an inequalityWorked Examples (system: \(y < 2x+1\) and \(y \ge -x+2\))
A. A point in the overlap
Is \((3, 0)\) a solution?
\(0 < 2(3)+1 = 7\) ✓ and \(0 \ge -3+2 = -1\) ✓. Both true → yes, it’s a solution.
B. A point that fails one
Is \((0, 0)\) a solution?
\(0 < 1\) ✓ but \(0 \ge 2\) ✗. One fails → not a solution.
C. Another solution
Is \((5, 1)\) a solution?
\(1 < 11\) ✓ and \(1 \ge -3\) ✓. Yes — it lies in the overlap.
D. Boundary matters
For \(y \ge -x + 2\), points on that line count (solid, inclusive); for \(y < 2x + 1\), points on that line do not (dashed, strict).
Solid includes its edge; dashed excludes it.
Where You’ll Use It
Systems of inequalities model real choices with multiple limits at once — a budget and a time cap, minimum protein and maximum calories, at least so many of one item and no more than another. The overlapping “feasible region” shows every combination that satisfies all the constraints, which is the foundation of linear programming.
Easy Points to Lose
- Shading only one inequality. The solution is the overlap; you must shade both and find where they meet.
- Wrong boundary style. Dashed for strict (\(<\), \(>\)); solid for inclusive (\(\le\), \(\ge\)).
- Calling a point a solution when it satisfies just one. It must make both inequalities true.
- Skipping the test point. Don’t guess the shading side — verify with \((0,0)\) (or another off-line point).
Your Turn (system: \(y < 2x+1\) and \(y \ge -x+2\))
Decide whether each point is a solution, then reveal the answers.
- \((4, 2)\)
- \((0, 3)\)
- \((5, 1)\)
- \((0, 0)\)
Show answers
- \(\color{blue}{\text{yes } (2<9 \text{ and } 2\ge -2)}\)
- \(\color{blue}{\text{no } (3<1 \text{ is false})}\)
- \(\color{blue}{\text{yes}}\)
- \(\color{blue}{\text{no } (0\ge 2 \text{ is false})}\)
Make Your Own Worksheet
Generate fresh systems-of-inequalities problems with a full answer key — print or save as a PDF.
Frequently Asked Questions
What is the solution to a system of linear inequalities?
It’s the region of the plane where the shaded areas of all the inequalities overlap. Every point in that region satisfies every inequality at once.
How do I know which side to shade?
Pick a test point off each boundary (usually the origin) and check it in that inequality. If it’s true, shade that side; if false, shade the other.
When is the boundary dashed vs. solid?
Dashed for strict inequalities (\(<\), \(>\)) because the line itself isn’t included; solid for \(\le\) and \(\ge\) because it is.
Can a system of inequalities have no solution?
Yes — if the two shaded regions don’t overlap anywhere (for example, parallel boundaries shaded away from each other), there’s no point that satisfies both.
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