Linear Inequalities in Two Variables · 6.2.3
Shading regions for inequality systems
Students draw the boundary line of each inequality in a system on the same axes, decide dashed or solid lines from the strict/non-strict sign, then shade the single region common to all the inequalities, the intersection representing every simultaneous solution.
The official learning standard (6.2.3)
“Determine and shade the region that satisfies a linear inequality system.”
What it means
Students draw the boundary line of each inequality in a system on the same axes, decide dashed or solid lines from the strict/non-strict sign, then shade the single region common to all the inequalities, the intersection representing every simultaneous solution.
How it is examined
A core Paper 2 skill: given two or three linear inequalities, students sketch all boundary lines on one Cartesian plane and shade the single region satisfying the whole system, usually earning marks for correct lines, correct shading, and clear labelling.
Worked example
Shade the region that satisfies x ≥ 0, y ≥ 0 and x + y ≤ 4.
- Draw x = 0 (the y-axis) and y = 0 (the x-axis) as solid boundary lines, since both inequalities are inclusive.
- Draw the line x + y = 4 as a solid line, passing through (4, 0) and (0, 4).
- x ≥ 0 gives the region on or right of the y-axis; y ≥ 0 gives the region on or above the x-axis; x + y ≤ 4 gives the region on or below the line.
- Shade the triangular region common to all three, bounded by (0, 0), (4, 0) and (0, 4).
Source:DSKP KSSM Mathematics Form 4 and 5 (Versi English)
Frequently asked questions
How do I know which side of a line to shade?
Substitute a test point not on the line, commonly (0, 0), into the inequality; if it makes the inequality true, shade the side containing that point, otherwise shade the opposite side.
Why do some lines need to be dashed?
A dashed line shows the boundary is not included because the inequality is strict (< or >), so points exactly on that line are not part of the solution.
What if the shaded regions from each inequality don't overlap?
Then the system would have no solution region, recheck each inequality and boundary line for errors, since typical exam questions are designed to always have a valid overlapping region.