The region for several inequalities is where all their conditions hold at once. Draw each boundary, find each side, and keep only the overlap.
This lesson is part of linear inequalities in two variables. It uses choosing the shaded side with a test point for each line.
What is the method?
- Draw every boundary, solid or dashed.
- Test a point for each line and decide its wanted side.
- Shade the unwanted side of each line, so the wanted region stays clear.
- Check one point inside the clear region against every inequality.
Worked example: three inequalities
An original problem: show the region R satisfying y ≥ 1, y ≤ 2x and x + y ≤ 6.
Boundaries. All three are solid.
y = 1 is horizontal. y = 2x passes through (0, 0) and (2, 4). x + y = 6 passes through (0, 6) and (6, 0).
Sides. For y ≥ 1, the wanted side is above the line. For y ≤ 2x, test (1, 0): 0 ≤ 2 is true, so the side containing (1, 0), below the line, is wanted.
For x + y ≤ 6, test (0, 0): 0 ≤ 6 is true, so the origin side is wanted.
Vertices. The region is a triangle. y = 1 meets y = 2x at (0.5, 1).
y = 1 meets x + y = 6 at (5, 1). y = 2x meets x + y = 6 where 3x = 6, so x = 2 and y = 4, giving (2, 4).
Check inside. Take (2, 2): 2 ≥ 1, 2 ≤ 4 and 4 ≤ 6 are all true.
Check outside. Take (1, 4): 4 ≥ 1 and 5 ≤ 6 are true, but 4 ≤ 2 is false, so (1, 4) is outside.
Which mistake makes a mess?
The common slip is shading the wanted side of every line and then trying to find the overlap by eye. On three lines the page becomes a crowd of shaded zones and the overlap is unclear.
Shading the unwanted side leaves a clean region. Label it R so your answer is unmistakable.
What changes with a dashed line?
Suppose the third inequality were x + y < 6. Then the line is dashed and the points on it, such as (2, 4), are not in the region. The region looks the same on the graph but the top edge is not included.
A check point on that edge now fails: 2 + 4 = 6, and 6 < 6 is false.
A self-check question
Show the region R for x ≥ 0, y ≥ 0 and x + y < 5. Is (2, 3) in R? Is (2, 2)?
Answer
The boundaries x = 0 and y = 0 are solid axes, and x + y = 5 is dashed. R is the triangle with corners (0, 0), (5, 0) and (0, 5), with the slanted edge not included.
For (2, 3): 2 + 3 = 5, and 5 < 5 is false, so it is not in R. For (2, 2): 4 < 5 is true, and both coordinates are at least 0, so it is in R.
Next, see translating constraints in a word problem. Log slips in the mistake log tool to see which step fails.
If multi-line graphs still feel crowded, SPM Mathematics one-to-one tuition lets a teacher show a layout that works on your own questions.