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MathChapter 4: Linear Inequalities and Graphs

Notes

Graphing Systems of Inequalities Notes

Find and interpret the intersection of multiple inequality regions using exact boundaries, candidate-point testing, and region analysis.

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A solution to a system must satisfy every inequality simultaneously. Each inequality contributes a half-plane; the final solution is their geometric intersection, not every individually shaded area.

Graph and intersect every constraint

System graphing workflow

  1. Graph the first boundary

    Determine its solid or dashed style and valid half-plane.

  2. Repeat

    Graph and shade every remaining inequality independently.

  3. Intersect

    Keep only points lying in every valid half-plane.

  4. Check boundaries

    A boundary segment belongs to the solution only when its inequality includes equality and all other constraints hold there.

  5. Verify a point

    Substitute one point from the final region into every original inequality.

Vertical, horizontal, and sloped constraints

Example system
\[\begin{cases}x\ge-3\y<4\y\ge\frac12x-1\end{cases}\]

The solution lies right of the vertical boundary, below the horizontal boundary, and above the sloped boundary.

Three-constraint solution regionCoordinate plane with solid vertical boundary x equals negative three, dashed horizontal boundary y equals four, and solid sloped boundary y equals one half x minus one. The highlighted overlap is right of the vertical line, below the horizontal line, and above the sloped line.-6-5-4-3-2-1123456-6-5-4-3-2-1123456xyx boundaryy boundarysloped boundary
Three-constraint solution region

The vertical and sloped boundaries are solid because their operators include equality. The horizontal boundary is dashed. The mint polygon is computed by clipping the coordinate window against all three algebraic half-planes.

Two sloped inequalities

Example system
\[\begin{cases}x+y\le3\x-2y<4\end{cases}\]

Equivalently, the solution is below or on \(y=-x+3\) and strictly above \(y=\frac12x-2\).

Intersection between two sloped boundariesCoordinate plane with solid boundary y equals negative x plus three and dashed boundary y equals one half x minus two. The overlap lies below the first line and above the second.-6-5-4-3-2-1123456-6-5-4-3-2-1123456xyupper boundarylower boundary
Intersection between two sloped boundaries

Test an ordered pair without graphing

Worked example

Test one candidate

Does \((2,1)\) satisfy \(x+y\le4\), \(y>x-3\), and \(x\ge0\)?

  1. First inequality

    \(2+1\le4\) is true.

  2. Second inequality

    \(1>2-3\) is true.

  3. Third inequality

    \(2\ge0\) is true.

  4. Conclude

    Every statement is true, so the point is a system solution.

Yes, \((2,1)\) satisfies all three constraints.

Identify a labeled solution region

Four regions divided by two linesTwo solid lines y equals x plus one and y equals negative x plus one divide the plane into four labeled regions. Region A, above both lines, is highlighted.-6-5-4-3-2-1123456-6-5-4-3-2-1123456xyABCDrising boundaryfalling boundary
Four regions divided by two lines

The highlighted Region A satisfies both \(y\ge x+1\) and \(y\ge-x+1\). A point in any other labeled region lies below at least one boundary and fails the system.

Quadrant analysis

To decide whether a quadrant contains solutions, combine sign restrictions with the system. For example, \(x\ge2\) and \(y\le-1\) force positive \(x\) and negative \(y\), so all solutions lie in Quadrant IV; the other three quadrants contain none.

Mini check

Check your understanding

Which point satisfies \(y\ge x-1\) and \(y<-x+5\)?

  1. \((0,0)\)
  2. \((2,2)\)
  3. \((4,0)\)
  4. \((3,5)\)
Show answer and explanation

Answer: \((2,2)\)

At \((2,2)\), \(2\ge1\) and \(2<3\). Each other point fails at least one inequality.

Key takeaways

What to remember

  • A system solution satisfies every inequality.
  • The final region is the intersection of all half-planes.
  • Boundary inclusion is decided separately for each inequality.
  • Candidate-point testing and labeled-region testing use the same all-statements-must-be-true rule.
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Apply the ideas with SAT-style questions, then reinforce key details with flashcards.