OxfordAQA International GCSE Mathematics: Inequalities on number lines and graphs (9260) – Revision Notes

Quick recall for reference A23 in section 3.2.3 of the OxfordAQA International GCSE Mathematics (9260) specification, Version 5.1, for exams from May/June 2018 onwards. The study guide has the full explanations and longer worked examples. To test yourself afterwards, try the practice questions, the course hub, the printable checklist and the free diagnostics.

Core content is also assessable at Extension. Parts marked Extension only are not Core.

Symbols and circles (Core)

Inequality	Boundary in the set?	Number-line circle	Graph line (Extension only)
x < a	No	Open ○	Dashed
x ≤ a	Yes	Closed ●	Solid
x > a	No	Open ○	Dashed
x ≥ a	Yes	Closed ●	Solid

x > −2                        x ≤ 3
○──────────────>              <──────────●
−3  −2  −1   0   1            1   2   3   4

−1 ≤ x < 4
  ●━━━━━━━━━━━━━━━○
 −2  −1  0  1  2  3  4  5

Reading from a line: the direction of shading gives the sign, and the circle gives strict or included.

Method for linear inequalities (Core)

1. Expand brackets and clear fractions (multiply by a positive number).
2. Gather the x terms together on one side, with the plain numbers on the other.
3. Divide both sides by the number multiplying x.
4. If you multiply or divide by a negative, reverse the sign.
5. Write the answer with x first, then draw the number line if asked.

Quick reminders:

Case	Working	Answer
Unknown both sides	5x + 4 > 2x + 19 → 3x > 15	x > 5
Negative coefficient	6 − 4x ≥ 18 → −4x ≥ 12	x ≤ −3
Brackets	2(x − 3) < 10 → x − 3 < 5	x < 8
Double inequality	−2 ≤ 3x + 1 < 13 → −3 ≤ 3x < 12	−1 ≤ x < 4

For a double inequality, apply every step to the left, middle and right parts.

For integer lists, check each end against its circle. The set −1 ≤ x < 4 contains −1, 0, 1, 2 and 3 only.

Two conditions together: find where both are true. For x < 6 and x ≥ −2 that is −2 ≤ x < 6. Draw both on separate lines and look for the overlap.

Quadratic inequalities (Extension only)

1. Rearrange to  ax² + bx + c  compared with 0.
2. Solve the matching equation (replace the sign with =)  → critical values.
3. Sketch the U-shape (a > 0) through the critical values.
4. "> 0": outside the roots.   "< 0": between the roots.

Form (roots p < q)	Solution set
(x − p)(x − q) > 0	x < p or x > q
(x − p)(x − q) ≥ 0	x ≤ p or x ≥ q
(x − p)(x − q) < 0	p < x < q
(x − p)(x − q) ≤ 0	p ≤ x ≤ q

Worked reminders:

x² − 7x + 10 < 0  →  (x − 2)(x − 5) < 0   →  2 < x < 5
x² ≥ 25           →  x² − 25 ≥ 0          →  x ≤ −5 or x ≥ 5
x² + x − 12 > 0   →  (x + 4)(x − 3) > 0   →  x < −4 or x > 3

If the x² coefficient is negative, multiply the whole inequality by −1 and reverse the sign first. For example, 12 − x − x² ≥ 0 becomes x² + x − 12 ≤ 0, so −4 ≤ x ≤ 3.

If the critical values are not whole numbers, use the formula or complete the square and keep surds in the answer when asked for an exact form.

Showing a two-part answer on a number line: draw one line with two separate pieces, each with an arrow pointing away from the middle. For x < −3 or x > 6, that is an open circle at −3 with an arrow heading left, and an open circle at 6 with an arrow heading right.

x < −3 or x > 6
<──────○                    ○──────>
−4  −3  −2  −1  0  1  2  3  4  5  6  7

Special cases (Extension only): a squared bracket is never negative.

Inequality	Solution
(x − 3)² ≥ 0	Every value of x
(x − 3)² > 0	Every x except x = 3
(x − 3)² ≤ 0	Only x = 3
(x − 3)² < 0	No solution

Regions on a graph (Extension only)

Method in steps:

1. Draw the boundary line for each inequality, treating the sign as =.
2. Solid for ≤ and ≥. Dashed for < and >.
3. Test a point not on the line (often (0, 0)).
4. True: shade that side. False: shade the other side.
5. The required region is where all the conditions hold. Check the question for whether to shade the wanted region or the unwanted parts.

Inequality	Region
y > mx + c	Above a dashed line
y ≤ mx + c	Below or on a solid line
x ≥ k	On or to the right of a solid vertical line
x < k	To the left of a dashed vertical line

To write inequalities from a graph: find each line’s equation from two points, check the line style, then test a point inside the region to choose the sign.

Must-know distinctions

• A strict inequality leaves out the boundary. An included inequality keeps it. Open circle and dashed line go with strict.
• 7 ≥ x and x ≤ 7 are the same statement.
• “And” means overlap of two ranges. “Or” means both ranges together.
• A quadratic “less than” gives one range. A quadratic “greater than” gives two ranges.
• Dividing by −2 flips the sign, but subtracting 2 does not.
• A solution set is a range of values, not a single number, so an answer such as x ≤ 4 describes infinitely many values unless the question asks for integers only.
• A boundary point can be in the set (closed circle, solid line) or out of it (open circle, dashed line). Always say which in your own words if you are unsure how to label a diagram.
• Critical values come from the equation. The inequality sign tells you which regions to keep.

Turning words into symbols (Core)

Words	Symbol
more than, greater than, over	>
less than, fewer than, under	<
at least, no less than, a minimum of	≥
at most, no more than, a maximum of	≤
between 3 and 8 (not including either)	3 < x < 8
from 3 to 8 inclusive	3 ≤ x ≤ 8

Checking an answer

Substitute one value from your solution set and one from outside it.

For 8x − 3 > 5x + 12 with answer x > 5:

x = 6:  8(6) − 3 = 45   and  5(6) + 12 = 42   → 45 > 42, true
x = 5:  8(5) − 3 = 37   and  5(5) + 12 = 37   → 37 > 37, false

The boundary value x = 5 gives equality, which is why the circle is open. If the question had used ≥, 5 would be in the set.

More worked reminders

Linear, with brackets and a negative coefficient:

4(3 − x) > 2x + 6
12 − 4x > 2x + 6
6 > 6x
1 > x        so   x < 1

Quadratic that needs rearranging first (Extension only):

x² < 6x − 8
x² − 6x + 8 < 0
(x − 2)(x − 4) < 0
2 < x < 4

Region with integer coordinates (Extension only): find the integer points satisfying y > x + 1, y ≤ 6 and x ≥ 0. Work column by column.

x	Allowed y (whole numbers)	Count
0	2, 3, 4, 5, 6	5
1	3, 4, 5, 6	4
2	4, 5, 6	3
3	5, 6	2
4	6	1

Total 15 points. A point on the dashed line y = x + 1, such as (2, 3), does not count.

Quick self-test

Cover the answers and try each question.

1. Solve 8x − 3 > 5x + 12.
2. Solve 13 − 2x ≤ 5.
3. Solve 3(x + 2) < 5x − 4.
4. List the integers satisfying −5 < 2x + 1 ≤ 9.
5. Write the inequality shown by a closed circle at −2, an open circle at 5 and a line between them.
6. Solve (x + 5)/4 ≥ 2 − x.
7. (Extension only) Solve x² − 5x − 14 < 0.
8. (Extension only) Solve x² > 64.
9. (Extension only) Solve 3x² + 5x − 2 ≥ 0.
10. (Extension only) A region lies on or above the line y = 3 and strictly below the line y = 2x − 2. Write the two inequalities and say how each line is drawn.
11. (Extension only) Solve x² ≤ 9.

Answers

1. 3x > 15, so x > 5.
2. −2x ≤ −8, so x ≥ 4.
3. 3x + 6 < 5x − 4, so 10 < 2x, so x > 5.
4. −6 < 2x ≤ 8, so −3 < x ≤ 4. The integers are −2, −1, 0, 1, 2, 3, 4.
5. −2 ≤ x < 5.
6. x + 5 ≥ 8 − 4x, so 5x ≥ 3, so x ≥ 3/5.
7. (x − 7)(x + 2) < 0, so −2 < x < 7.
8. x² − 64 > 0, so x < −8 or x > 8.
9. (3x − 1)(x + 2) ≥ 0, so x ≤ −2 or x ≥ 1/3.
10. y ≥ 3 with a solid line, and y < 2x − 2 with a dashed line.
11. x² − 9 ≤ 0, so −3 ≤ x ≤ 3.

Where marks are usually lost

• Multiplying or dividing through by a negative and leaving the sign as it was.
• Giving a final answer as an equation, such as x = 4, when the question asks for an inequality.
• Open and closed circles swapped on a number line.
• Leaving a quadratic answer as a single range, such as x > 5, when two ranges are needed.
• Writing a double-ended range for “outside the roots”, such as −3 > x > 6.
• Dropping surds into rounded decimals when the question asks for an exact answer.
• A dashed line where the inequality includes the boundary, or the reverse.
• Counting integers on an open end, or missing one at a closed end.
• Shading without a test point and ending up on the wrong side.

Official syllabus

OxfordAQA International GCSE Mathematics (9260), Version 5.1 (for exams May/June 2018 onwards), published by OxfordAQA. Section 3.2.3, reference A23.