Inequality regions and boundary lines
Use half-planes to show all points that satisfy several linear inequalities.
Extended only.
Before you begin
- Plot coordinates and draw a straight line from two points.
- Interpret strict and inclusive inequalities.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which equation represents a vertical line through x = 2?
What you will learn
- Draw solid or broken inequality boundaries.
- Use a test point to identify the permitted half-plane.
- Describe a region with simultaneous inequalities.
Draw the equality as the boundary
For y > 2x + 1, first draw the line y = 2x + 1 through points such as (0, 1) and (1, 3). Use a broken line because the equality is excluded. For y ≥ 2x + 1, use a solid line because the boundary points are allowed. Horizontal and vertical inequalities use y = k and x = k as their boundaries.
Substitute a convenient point that is not on the boundary to decide which half-plane is permitted. For (0, 0), the inequality 0 > 1 is false, so the permitted side is the side opposite the origin. A point on the boundary cannot distinguish the two half-planes.
Show the unwanted region according to the question
In the Cambridge convention, shade the unwanted half-plane unless the question directs otherwise. For y ≥ 2x + 1, shade below the line and leave the region above unshaded; the solid boundary remains permitted. State which convention you are following so a shaded graph is not misread.
For x ≥ 0, y ≥ 0 and x + y ≤ 6, the allowed region lies in the first quadrant below the solid line x + y = 6. The three boundary intersections form a triangle with vertices (0, 0), (6, 0) and (0, 6). The region includes interior points as well as its boundaries.
Read a region back into inequalities
If a region is to the right of x = 1, above y = 2 and below x + y = 7, its inequalities are x ≥ 1, y ≥ 2 and x + y ≤ 7 when all lines are solid. Change an inclusive sign to a strict sign when its boundary is broken. Test an interior point such as (2, 3) against every condition.
A pair must satisfy every inequality to belong to the common region. The fact that it satisfies two of three is insufficient. If a problem restricts the coordinates to integers, check those discrete points after drawing the continuous region. This lesson describes regions; it does not introduce linear-programming optimisation.
Pause and explain
For y < x + 1, which boundary and side are permitted?
Worked example
Describe the region x ≥ 0, y ≥ 0, x + y ≤ 6, and decide whether (2, 4) and (3, 4) belong.
Show the worked solution
- Draw both axes and the line joining (6, 0) to (0, 6), all solid.
- The permitted region is the first-quadrant triangle on or below x + y = 6; shade outside it.
- (2, 4) is on the included boundary. (3, 4) fails because 3 + 4 = 7 > 6.
Answer Triangle with vertices (0, 0), (6, 0), (0, 6); (2, 4) belongs, (3, 4) does 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 the region x < 3 and y ≥ 1, using unwanted-region shading.
Give me a hint
The two boundaries are one vertical and one horizontal line.
Compare my reasoning
- Use a broken vertical line x = 3 and a solid horizontal line y = 1.
- Allowed points are left of x = 3 and on or above y = 1.
- Shade the right and lower unwanted regions; (2, 2) checks the remaining overlap.
Left of x = 3, on or above y = 1; x boundary excluded
Look for these in your work
- Chose each line style from the sign.
- Left the common allowed region unshaded.
A region is right of solid x = 1, above broken y = 2, and below solid x + y = 8. Write its inequalities.
Give me a hint
Solid includes equality; broken excludes it.
Compare my reasoning
- The right-hand condition is x ≥ 1.
- The height condition is y > 2.
- The upper boundary gives x + y ≤ 8; all three conditions must hold.
x ≥ 1, y > 2, x + y ≤ 8
Look for these in your work
- Interpreted the permitted side of each boundary.
- Recorded strictness independently for each line.
A learner chooses non-negative integer counts x of maths tasks and y of reading tasks. A session allows at most six tasks and requires at least two reading tasks. Describe the region and list possible y when x = 3.
Give me a hint
Combine the total limit with positivity and the reading minimum.
Compare my reasoning
- The constraints are x ≥ 0, y ≥ 2 and x + y ≤ 6, with integer counts.
- At x = 3, the total limit gives y ≤ 3.
- Combined with y ≥ 2, the choices are y = 2 or 3.
x ≥ 0, y ≥ 2, x + y ≤ 6; at x = 3, y = 2 or 3
Look for these in your work
- Translated all three requirements.
- Applied the integer restriction after intersecting the conditions.
Common mistakes
- Using a solid line for a strict condition.
- Shading the allowed side when the question asks for unwanted-region shading.
- Checking only one inequality for a point.
What can you explain now?
Does (2, 3) satisfy x > 1, y ≥ 2 and x + y < 5?
Compare with the explanation
No
The first two conditions hold, but 2 + 3 = 5 is excluded by the final strict inequality.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Draw a region with one strict boundary. Give a point inside, outside and on that boundary, then justify whether each is allowed.
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.