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.
Solve Systems of Linear Inequalities: what to notice and how to work it
What to notice first
Common student mistake
Key formulas and cues
A reliable path
- Solve the boundaryTemporarily treat the inequality like an equation.
- Choose the sideUse the sign or test a number if the direction is not obvious.
- Graph the solutionUse the correct endpoint and shade the values that work.
Worked examples
Flip the sign
- Divide both sides by -3.
- Reverse the inequality sign.
- Simplify 12 divided by -3.
Keep the sign
- Subtract 5 from both sides.
- No negative multiplication or division happened.
- Keep the sign direction.
Try one before moving on
Solve Systems of Linear Inequalities: pop-up practice
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
How do you solve systems of linear inequalities by graphing?
Graph each inequality’s boundary line, using a solid line for ≤ or ≥ and a dashed line for < or >. Shade the side of each line that makes the inequality true; a test point such as (0, 0) tells you which side. The solution is the region where the shadings overlap. For y < 2x + 1 and y ≥ −x + 2, the overlap lies to the right of where the lines cross at (1/3, 5/3).
How do you know if a point is a solution to a system of linear inequalities?
Substitute the point’s x and y values into every inequality. It is a solution only if all of them come out true. Check (2, 1) in the system y < 2x + 1 and y ≥ −x + 2: 1 < 5 is true and 1 ≥ 0 is true, so (2, 1) works. The point (0, 0) fails, because 0 ≥ 2 is false, even though 0 < 1 is true.
Can you solve systems of linear inequalities without graphing?
Partly. You can find the corner points of the solution region algebraically by solving the boundary equations together, and you can test any point by substitution. For a one-variable system, algebra is enough: 2x + 1 > 5 and x − 3 ≤ 4 give x > 2 and x ≤ 7, so 2 < x ≤ 7. In two variables, though, the full answer is a region, so a graph is the clearest way to show it.
Can a system of linear inequalities have exactly one solution?
Usually not. Most systems have infinitely many solutions (a shaded region) or none at all, such as y > 2x + 3 and y < 2x − 1, whose parallel boundaries never let the shadings meet. A single point is possible only in special cases with ≤ and ≥ pinned together. For example, x ≥ 3, x ≤ 3, y ≥ 1 and y ≤ 1 allow only the point (3, 1).
Is solving inequalities the same as solving equations?
Almost. You use the same steps, such as adding, subtracting, multiplying and dividing both sides, but with one key difference: multiplying or dividing by a negative number reverses the inequality sign. For example, −2x ≤ 10 becomes x ≥ −5 after dividing by −2. Also, an equation usually has one answer, while an inequality has a whole set of answers shown with shading.
Is there a solving systems of linear inequalities worksheet PDF?
Yes. Printable worksheets on systems of linear inequalities usually ask students to graph two boundary lines, shade, and name a point in the solution region. When practicing, label each line with its inequality and choose a test point that is not on either line. Checking one point from the overlap in both inequalities is a quick way to confirm the shading is correct.
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