Inequalities: Number Lines, Solving and Graphing Regions
Core’s entire syllabus requirement here is reading and representing inequalities on a number line — already touched on back in the Number unit. Everything beyond that — solving inequalities algebraically, graphing them as regions, and listing the inequalities that define a shape — is Extended-only, tagged throughout.
1. Number Line Notation
A quick recap: open circles mark strict inequalities (<, >) because the boundary value itself isn’t included; closed (filled) circles mark inclusive inequalities (⩽, ⩾) because it is.
- Write down the inequality shown by a number line with a filled circle at 2 and an open circle at 6, shaded between them.
- Describe how you would draw −4 < x ⩽ 0 on a number line.
- List all integer values of x that satisfy −2 ⩽ x < 3.
Show answers
- 2 ⩽ x < 6
- Open circle at −4 (not included), filled circle at 0 (included), shaded between them.
- −2, −1, 0, 1, 2
2. Constructing and Solving Linear Inequalities Extended only
Solve exactly like a linear equation — with one crucial extra rule: flip the inequality sign whenever you multiply or divide both sides by a negative number.
3x − 2x < 4 → x < 4
−2x > 4
Divide both sides by −2, and flip the inequality: x < −2
3 + 6 ⩾ 7x − 4x → 9 ⩾ 3x
Divide by 3 (positive, no flip needed): 3 ⩾ x, i.e. x ⩽ 3
- Solve 5x + 2 > 17.
- Solve 8 − 3x ⩽ 2.
- Solve 6x − 5 < 3x + 10.
- Solve 2(x − 3) ⩾ x + 1.
Show answers
- 5x > 15 → x > 3
- −3x ⩽ −6 → divide by −3, flip: x ⩾ 2
- 3x < 15 → x < 5
- 2x − 6 ⩾ x + 1 → x ⩾ 7
3. Compound (Double) Inequalities Extended only
When a value is trapped between two limits, solve by applying the same operation to all three parts at once.
Add 2 to all three parts: −1 ⩽ 3x < 9
Divide all three parts by 3: −1/3 ⩽ x < 3
- Solve −4 ⩽ 2x + 2 < 10.
- Solve 1 < 4x − 3 ⩽ 13.
- Solve −6 ⩽ 5 − x < 2 (careful — this one needs a sign flip somewhere).
Show answers
- Subtract 2: −6 ⩽ 2x < 8; divide by 2: −3 ⩽ x < 4
- Add 3: 4 < 4x ⩽ 16; divide by 4: 1 < x ⩽ 4
- Subtract 5: −11 ⩽ −x < −3; divide by −1 and flip both inequalities (and reverse the order): 3 < x ⩽ 11
4. Graphing Linear Inequalities in Two Variables Extended only
Conventions: a broken line for a strict inequality (<, >), a solid line for an inclusive one (⩽, ⩾), and shading over the unwanted region, leaving the wanted region clear — unless a question tells you otherwise.
The test point method
Not sure which side of a line to shade? Pick any point not on the line — the origin (0, 0) is usually easiest — and substitute it into the inequality. If it’s true, that point’s side is the wanted region; if false, shade that side.
Test the origin in x + y ⩽ 6: 0 + 0 = 0 ⩽ 6 — true, so the origin’s side of the line x + y = 6 is the wanted side. The shaded triangle here is the wanted region (highlighted for clarity in this diagram) — bounded by the y-axis (x ⩾ 0), the x-axis (y ⩾ 0), and the solid line x + y = 6. On your own exam diagrams, remember the convention is the other way round: shade the unwanted area outside the region, unless the question says otherwise.
- Describe the line style and shading needed to show x ⩽ 4.
- Describe the line style and shading needed to show y > −2.
- Use the test point (0,0) to decide which side of the line x + y = 3 satisfies x + y < 3.
Show answers
- A solid vertical line at x = 4, with the unwanted region (x > 4) shaded.
- A broken horizontal line at y = −2, with the unwanted region (y ⩽ −2) shaded.
- At (0,0): 0 + 0 = 0 < 3 — true, so the origin's side (the side containing the origin) satisfies the inequality.
5. Listing Inequalities That Define a Region Extended only
This is the reverse of graphing — given a shaded/highlighted region and its boundary lines, write down every inequality needed to describe it exactly, including whether each boundary is included (solid) or not (broken).
The region sits above the x-axis: y ⩾ 0
The region sits to the right of the y-axis: x ⩾ 0
The region sits on the origin’s side of the solid line x + y = 6: x + y ⩽ 6
- A region is bounded by a solid vertical line at x = 5 (region to the left), a solid horizontal line at y = 2 (region above), and the origin is not in the region. Write down the two inequalities.
- A triangular region is bounded by x = 0, y = 0, and the broken line x + y = 4, with the region below/left of that line (not including the line itself). Write down the three inequalities.
Show answers
- x ⩽ 5 and y ⩾ 2
- x ⩾ 0, y ⩾ 0, and x + y < 4
- Forgetting to flip the inequality sign when multiplying or dividing both sides by a negative number.
- Using a solid line for a strict inequality, or a broken line for an inclusive one — check the symbol carefully (⩽/⩾ = solid, </> = broken).
- Shading the wanted region instead of the unwanted region (or vice versa, if a question specifically asks to shade the wanted region — always re-read what’s being asked).
- In a compound inequality, only applying an operation to one part instead of all three.
- Reversing the direction of a compound inequality incorrectly after multiplying/dividing by a negative — both inequality symbols must flip, and the order of the expression effectively reverses too.
6. Quick-Fire Challenge Round
- An open circle on a number line means the value is included.
- Dividing both sides of an inequality by −4 requires flipping the inequality sign.
- A strict inequality (< or >) is drawn with a solid line on a graph.
- The origin is always a safe choice for the test point method.
Show answers
- False — an open circle means the value is not included; a filled circle means it is.
- True.
- False — strict inequalities use a broken (dashed) line; inclusive inequalities use a solid line.
- False — it’s usually the easiest choice, but only if the line doesn’t pass through the origin itself; otherwise pick a different point.
- “Solve −3x > 12: dividing both sides by −3 gives x > −4.”
- “To show y ⩽ 2 on a graph, I drew a broken line at y = 2.”
- “Solve −5 ⩽ 2x < 9: dividing by 2 (positive, no flip) gives −5 ⩽ x < 9.”
Show answers
- Dividing by a negative number requires flipping the inequality: −3x > 12 → x < −4, not x > −4.
- ⩽ is an inclusive inequality, so it needs a solid line, not a broken one.
- Every part of the compound inequality must be divided by 2, including both bounds: −5/2 ⩽ x < 9/2, not −5 ⩽ x < 9 (the bounds weren't actually divided).
7. Mixed Practice — Bringing It All Together
- List the integers that satisfy −5 < x ⩽ 0.
- Draw a description of how −1 ⩽ x < 4 would look on a number line.
Show answers
- −4, −3, −2, −1, 0
- Filled circle at −1, open circle at 4, shaded between them.
- Solve 7 − 2x ⩽ 1.
- Solve −2 < 4x + 6 ⩽ 18.
- Describe the line style and shaded region for x + y > 4.
- Write down the inequalities defining a region bounded by a solid line at x = 0 (region to the right), a broken line at y = 5 (region below), and the x-axis (region above, solid).
Show answers
- −2x ⩽ −6 → x ⩾ 3
- Subtract 6: −8 < 4x ⩽ 12; divide by 4: −2 < x ⩽ 3
- A broken line at x + y = 4, with the region containing the origin (x + y < 4, the “unwanted” side per convention) shaded, leaving x + y > 4 clear.
- x ⩾ 0, y < 5, y ⩾ 0
Practice at Home
A large set again — work through it steadily.
- List the integers that satisfy −3 ⩽ x < 2.
- Describe how 0 < x ⩽ 5 would look on a number line.
- Write down the inequality shown by a number line with an open circle at −2 and a filled circle at 3, shaded between.
- Solve 4x − 7 > 9.
- Solve 10 − 3x ⩾ 1.
- Solve 5x + 4 < 2x − 5.
- Solve 3(x + 2) ⩽ 2(x + 8).
- Solve −6 ⩽ 3x + 3 < 15.
- Solve 2 < 5x − 3 ⩽ 17.
- Solve −4 ⩽ 8 − 2x < 4.
- Describe the line style and shading for y ⩾ −1.
- Describe the line style and shading for x < 3.
- Use the test point (0,0) to determine which side of x + 2y = 8 satisfies x + 2y < 8.
- A region is bounded by x = 0 (solid, region to the right), y = 0 (solid, region above), and x + y = 8 (solid, region containing the origin). Write down the three inequalities.
- A number x satisfies both 2x + 1 > 9 and 3x − 4 < 20. Find the range of possible values of x.
- A school hall can seat at most 250 people. If s is the number of students and t is the number of teachers, write an inequality connecting s and t.
- Explain, using an example, why the inequality sign must flip when both sides of an inequality are multiplied by a negative number, but not when both sides are multiplied by a positive number.
Key Vocabulary Recap
Syllabus Reference
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