Draw the intersection of linear inequality regions
Draw boundary lines with strict or inclusive styles · Select the correct half-plane using a test point · Find the intersection and check boundary membership
Section Paper 4 · Historical Extended · calculator
Exam scope and assessment notes
Extended content; not a Core requirement. Historical study of Cambridge IGCSE Mathematics (US) 0444, 2025 Version 2. The final assessment was June 2025; the syllabus was available only to schools in Arizona, USA. This is not a current exam-entry option. Original self-check teaching and explained practice are not Cambridge papers, official marks or predicted grades. Linked teaching foundations do not establish full syllabus mastery. Calibrated cumulative assessment, reliable drawn-work review and qualified mathematics-teacher review remain unfinished.
Before you begin
- Line equations
- Linear inequalities
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What should you establish before applying the method in this lesson?
What you will learn
- Draw boundary lines with strict or inclusive styles
- Select the correct half-plane using a test point
- Find the intersection and check boundary membership
Line first, inequality second
Replace a linear inequality by equality to draw its boundary. Use a solid line for ≤ or ≥, and a dashed line for < or >. Choose a point not on the boundary and test it in the original inequality to determine the accepted side.
For y ≥ 2x − 1 the boundary is y = 2x − 1, with the region on or above it. Rearranging inequalities needs care when dividing by a negative coefficient: reverse the inequality sign.
Every condition must hold
For several inequalities, retain only their intersection, the points satisfying all conditions simultaneously. Boundary intersections help locate the region, but a corner involving a strict inequality is not itself included.
Validate an interior point and a boundary point in all original inequalities. A drawing's window may show only part of an unbounded region; do not infer a bound that was never specified.
Sources for this lesson
Pause and explain
How should the boundary of y < 2x + 1 be drawn?
Worked example
Describe y > 1, y ≤ x and x + y < 6. Are (2, 2) and (3, 3) included?
Show the worked solution
- y = 1 and x + y = 6 are dashed boundaries; y = x is solid.
- (2, 2) gives 2 > 1, 2 ≤ 2 and 4 < 6: included.
- (3, 3) gives 6 < 6, which is false: excluded. The boundary vertices (1, 1), (5, 1), (3, 3) are all excluded by a strict condition.
Answer Keep the common region above y = 1, on/below y = x and below x + y = 6; (2, 2) is included, (3, 3) is not.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Draw y ≥ 2x − 1. Which side contains (0, 0), and is (1, 1) included?
Give me a hint
Test both points in the inequality.
Compare my reasoning
- Boundary is solid. At (0, 0), 0 ≥ −1 is true, so shade the side containing the origin.
- At (1, 1), 1 ≥ 1 is true; it lies on the accepted boundary.
Solid boundary; shade on/above it. Both points are included.
Look for these in your work
- Use the original condition
- Allow equality
For x ≥ 0, y ≥ 0, x + y < 4, check (1, 2), (2, 2) and (−1, 2). Describe the boundaries.
Give me a hint
Check all three conditions per point.
Compare my reasoning
- (1, 2) satisfies each; (2, 2) fails the strict sum; (−1, 2) fails x ≥ 0.
- Axes are solid; x + y = 4 is dashed.
Only (1, 2) is included. Keep the first-quadrant region below the dashed line x + y = 4.
Look for these in your work
- Intersect rather than combine accepted sides loosely
- Reject strict-boundary points
A workshop makes x tables and y chairs, with x ≥ 0, y ≥ 0 and 2x + y ≤ 6. Describe the continuous region, then list feasible whole-number y values when x = 2.
Give me a hint
A count context adds integer restrictions to the shaded region.
Compare my reasoning
- The solid resource boundary joins (0, 6) and (3, 0); the accepted triangle includes its boundaries.
- At x = 2, y ≤ 2 and y ≥ 0. Whole-number counts are 0, 1, 2.
Closed triangle with vertices (0, 0), (0, 6), (3, 0). For x = 2, feasible counts y = 0, 1, 2.
Look for these in your work
- Include the resource boundary
- Distinguish continuous shading from count feasibility
Common mistakes
- A point on one accepted boundary can still fail another condition.
- Strict inequalities exclude their boundary, even where lines intersect.
What can you explain now?
How should the boundary of y < 2x + 1 be drawn?
Compare with the explanation
Dashed, with the accepted region below it.
Equality is excluded by the strict inequality.
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Make it stick
How should the boundary of y < 2x + 1 be drawn?
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.