Linear Inequalities in Two Variables · Form 4
Linear Inequalities in Two Variables: Common Mistakes
The mistakes that quietly cost marks in Linear Inequalities in Two Variables, and how to avoid each one in the SPM exam.
In our experience teaching Linear Inequalities in Two Variables, most lost marks come from a small set of repeated slips, not from a lack of understanding. Here they are, with the fix for each.
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
Six more slips that quietly cost region marks
- What students write: from −2y > 4x they get y > −2x. → Why it loses marks: dividing an inequality by a negative number reverses the sign, so the region drawn ends up on the exact opposite side. → Correct working: divide by −2 AND flip the sign: y < −2x.
- What students write: they test (0, 0) to pick a side even though the boundary passes through the origin. → Why it loses marks: a point on the line gives 0 < 0 or 0 = 0, which decides nothing, so the shading becomes a guess. → Correct working: choose a point clearly off the line, such as (1, 0) or (0, 1), and test that instead.
- What students write: they shade every region touched by any single inequality. → Why it loses marks: the required answer is the region satisfying ALL inequalities at once, the overlap, not the combined area. → Correct working: shade only where all the shadings overlap, then label that overlap R.
- What students write: 'at least 30' becomes x < 30. → Why it loses marks: 'at least' means 30 or more, so the symbol must be ≥, not <; the whole region then sits on the wrong side. → Correct working: at least → ≥, at most → ≤, more than → >, less than → <; here write x ≥ 30.
- What students write: they plot y = 2x + 4 from the y-intercept alone and guess the slant. → Why it loses marks: one point cannot fix a line, so the boundary is drawn at the wrong angle and the region is wrong. → Correct working: take two points, at x = 0, y = 4; at y = 0, x = −2, plot (0, 4) and (−2, 0), then rule the line.
- What students write: for x + y < 6 they list (3, 3) as part of the region. → Why it loses marks: the inequality is strict (<), so points on the boundary x + y = 6 are excluded, and (3, 3) gives 6, not less than 6. → Correct working: (3, 3) lies ON the dashed line, so it is NOT in the region; pick a point where x + y is strictly below 6.
The single costliest slip
Source:DSKP KSSM Mathematics Form 4 and 5 (Versi English)
Frequently asked questions
I rearranged the inequality and I'm unsure if the sign should have flipped, how do I check?
Ask one question: did I multiply or divide both sides by a negative number? If yes, the sign flips; if no, it stays.
Then test a number: pick any easy value that clearly satisfies the original, and confirm it still satisfies your rearranged version. If it does not, your sign is wrong.
My tested point gave a true statement but the answer says I shaded the wrong side, why?
Almost always the test point sits on the boundary line, not off it, so its true statement proves nothing about a side. Re-check by choosing a point that is plainly inside one region and clearly not on any line.
The origin only works when no boundary passes through (0, 0).
Do I lose marks if my region is correct but I used the opposite shading convention?
No, as long as the region you mean is unambiguous. Shading the wanted area or shading everything except it are both accepted, provided you add a key and clearly label the final region R.
Marks are lost only when the marker cannot tell which area is your answer, so make the labelling explicit.