Form 4 · Relationship and Algebra
Linear Inequalities in Two Variables
Linear inequalities describe regions, not single answers, the maths behind “at most”, “at least” and shaded graphs.
What is Linear Inequalities in Two Variables?
Instead of a line, an inequality like y ≥ 2x + 1 describes a whole region of the graph. This chapter teaches you to draw that region, combine several inequalities into one feasible area, and read real constraints, budgets, limits, minimums, into mathematical form.
Content standards (DSKP)
The DSKP KSSM sets these content standards for this chapter:
- 6.1 Linear Inequalities in Two Variables
- 6.2 Systems of Linear Inequalities in Two Variables
The key ideas
Solid vs dashed boundary
Use a solid line for ≤ or ≥ (the boundary is included) and a dashed line for < or > (it is not).
Which side to shade
Test a point not on the line (often the origin) to decide which side satisfies the inequality.
The feasible region
When several inequalities are shaded together, the region satisfying all of them is where the answer lives.
How this chapter is examined
Paper 2 typically asks you to draw two or three inequalities and identify the region satisfying all of them, sometimes from a real constraint like a shop’s stock limits. Neat lines and a clearly labelled region earn the marks; a rushed graph loses them.
How to study this chapter
Common mistakes to avoid
- Using a solid line when the inequality is strict (< or >)
- Shading the wrong side of a boundary
- Forgetting a hidden constraint like x ≥ 0 in a real problem
The test-point method: stop guessing which side
Once you have drawn a boundary line, deciding which side satisfies the inequality does not need any intuition, one test point settles it. Pick a point that is clearly on one side and not on the line itself; the origin (0, 0) is easiest because the arithmetic is trivial, as long as the line does not pass through it.
Substitute the point's coordinates into the inequality. If the statement comes out true, shade the side that contains your test point; if it comes out false, shade the other side.
For example, to place y ≥ 2x + 1, test (0, 0): is 0 ≥ 2(0) + 1, that is 0 ≥ 1? That is false, so you shade the side away from the origin.
If the boundary does pass through the origin, just choose another easy point such as (1, 0).
Turning words into inequalities
Real-context questions hide the inequalities inside sentences, and the marks start with translating them correctly. Learn the standard phrases: 'at least' and 'not less than' mean ≥; 'at most' and 'not more than' mean ≤; 'more than' means the strict >; 'fewer than' or 'less than' means the strict <.
Watch for comparisons between the two variables too 'the number of chairs is at least twice the number of tables' becomes y ≥ 2x if y is chairs and x is tables. Finally, many contexts carry hidden non-negativity: you cannot have a negative number of items, hours or ringgit, so add x ≥ 0 and y ≥ 0 even when the words never say so.
Missing these silent constraints is one of the most common ways to lose a mark on an otherwise correct region.
Confirm your region with one point inside it
The feasible region is the overlap where every inequality is true at once, so the fastest way to check your final shading is to pick any point that sits inside your shaded area and test it against all of the inequalities in turn. If it satisfies every one, your region is in the right place; if even one fails, the shading is wrong somewhere and you should re-check that particular boundary.
Choose a point well away from the edges so a near-miss does not fool you, the middle of the region is ideal. This one habit catches the usual errors (a line shaded on the wrong side, or a constraint forgotten entirely) before you lose the marks, and it costs only a few seconds of substitution.
A worked exam-style example
This example works through defining a region, finding where two boundaries meet, and testing a point, the core skills of the chapter.
- (a) Each inequality includes equality (≤ or ≥), so every point on the line counts and all three boundary lines are drawn solid.
- (b) To find where y = 2x meets x + y = 12, substitute y = 2x into the second equation: x + 2x = 12.
- So 3x = 12, giving x = 4, and then y = 2x = 2(4) = 8.
- The intersection point is (4, 8).
- (c) Test (5, 4) against each inequality in turn.
- First, y ≤ 2x: is 4 ≤ 2(5) = 10? Yes, 4 ≤ 10 is true.
- Second, x + y ≤ 12: is 5 + 4 = 9 ≤ 12? Yes, true.
- Third, y ≥ 2: is 4 ≥ 2? Yes, true.
- All three inequalities are satisfied, so (5, 4) lies inside R.
Study Linear Inequalities in Two Variables
Frequently asked questions
How this chapter is examined
SPM Mathematics assesses this chapter across Mathematics Paper 1 (Objective) and Mathematics Paper 2 (Subjective), drawing on the DSKP content standards above. Paper 2 gives marks for working, so showing every step matters.
Common mistakes to avoid
Using a solid line when the inequality is strict (< or >); Shading the wrong side of a boundary; Forgetting a hidden constraint like x ≥ 0 in a real problem.
Are any Linear Inequalities in Two Variables formulae given in the exam?
This chapter has no formula on the exam formula sheet, the working is expected from memory and method.
Solid or dashed line, how do I remember which?
If the inequality includes equality, the symbols ≤ or ≥, the points on the line count, so draw a solid line. If it is strict, < or > the line itself is excluded, so draw it dashed.
A memory hook: the small bar underneath ≤ and ≥ becomes the solid, unbroken line on your graph.
How do I decide which side of the line to shade?
Pick a test point that is not on the line, (0, 0) is easiest when the line misses the origin. Put its coordinates into the inequality.
If the statement is true, shade the side containing that point; if it is false, shade the other side. One substitution settles it every time, with no guessing.
The question says x and y are positive but gives no inequality for it, what do I do?
Add x ≥ 0 and y ≥ 0 yourself, they are hidden constraints. Many real contexts, like numbers of items, hours or ringgit, cannot be negative, so the region stays in the first quadrant even when the words only imply it.
Forgetting these silent conditions is a common way to lose an easy mark.
Source:DSKP KSSM Mathematics Form 4 and 5 (Versi English)· SPM: Format Pentaksiran mulai 2021, Matematik (1449)
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