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

This study guide covers reference A23 in section 3.2.3 (Solving equations and inequalities) of the OxfordAQA International GCSE Mathematics (9260) specification, Version 5.1, for exams from May/June 2018 onwards. The specification lists Core content and extra Extension content. Core content can also be assessed on the Extension papers, so you need it at both levels. The Extension-only parts are labelled Extension only below.

The rest of section 3.2.3 has its own pages: linear equations and forming equations and solving quadratic equations. Drawing straight lines is in the coordinates and straight-line graphs guide. Use the course hub and the printable checklist to track progress, and take a free diagnostic to find weak spots.

Also in this set: the revision notes and the practice questions.

What A23 asks for

Skill	Level
Solve linear inequalities in one variable	Core
Show the solution set on a number line, with an open circle for a strict inequality and a closed circle for an included boundary	Core
Solve linear inequalities in two variables	Extension only
Solve quadratic inequalities in one variable	Extension only
Show a solution set on a graph, using a dashed line for a strict inequality and a solid line for an included boundary	Extension only

Inequality symbols

Symbol	Meaning	Boundary included?
x < 3	x is less than 3	No (strict)
x ≤ 3	x is less than or equal to 3	Yes
x > 3	x is greater than 3	No (strict)
x ≥ 3	x is greater than or equal to 3	Yes

A double inequality such as −1 < x ≤ 4 means both parts are true at once: x is greater than −1 and also less than or equal to 4.

Number lines (Core)

A solution set is every value that makes the inequality true. Draw it on a number line like this:

• an open circle (○) where the boundary is not included (< or >)
• a closed circle (●) where the boundary is included (≤ or ≥)
• a line or arrow over the values that work

x ≤ 4                      −1 < x ≤ 4
<──────────────●           ──────○━━━━━━━━━━●──────
0  1  2  3  4  5           −2  −1  0  1  2  3  4  5
(arrow points left)        (open at −1, closed at 4)

Reading a line back into symbols: look at which way the shading goes, then at each circle. Shading to the right of a closed circle at 2 is x ≥ 2. Shading between an open circle at −3 and a closed circle at 1 is −3 < x ≤ 1.

Solving linear inequalities (Core)

Solve exactly as you would an equation, with one extra rule.

Multiplying or dividing by a negative number reverses the inequality sign. Adding or subtracting any number never changes it.

Why: 2 < 5 is true, but −2 and −5 swap order, so −2 > −5. Multiplying by −1 flips the sign.

Worked example 1: unknown on both sides

Solve 7x − 5 ≤ 3x + 11 and show the solution on a number line.

7x − 5 ≤ 3x + 11
7x − 3x ≤ 11 + 5      (collect x terms on one side)
4x ≤ 16
x ≤ 4                  (divide by 4, positive, so no change)

Closed circle at 4, with the line or arrow going left.

Worked example 2: negative coefficient

Solve 9 − 2x > 15.

9 − 2x > 15
−2x > 6               (subtract 9 from both sides)
x < −3                (divide by −2, so reverse the sign)

Open circle at −3, arrow going left.

A safer route avoids dividing by a negative: add 2x to both sides to get 9 > 15 + 2x, then 2x < −6, so x < −3. You get the same answer.

Worked example 3: fractions

Solve (2x + 1)/3 ≥ x − 2.

2x + 1 ≥ 3(x − 2)     (multiply both sides by 3, positive)
2x + 1 ≥ 3x − 6
7 ≥ x                 (add 6, subtract 2x)
x ≤ 7                 (read it back the right way round)

When the x term ends up on the right, rewrite the answer with x first. 7 ≥ x says exactly the same as x ≤ 7.

Worked example 4: double inequality

Solve −7 < 4x − 3 ≤ 13 and list the integer values.

−7 < 4x − 3 ≤ 13
−4 < 4x ≤ 16          (add 3 to all three parts)
−1 < x ≤ 4            (divide all three parts by 4)

The integers are 0, 1, 2, 3, 4. The integer −1 is not included because the circle at −1 is open.

Worked example 5: two conditions together

List the integers that satisfy both 2x + 3 < 15 and x ≥ −2.

The first inequality gives 2x < 12, so x < 6. Together with x ≥ −2 this is −2 ≤ x < 6. The integers are −2, −1, 0, 1, 2, 3, 4, 5.

Quadratic inequalities in one variable (Extension only)

The method uses the shape of a quadratic graph.

1. Rearrange so that one side is zero.
2. Solve the related equation to find the critical values. Factorise where possible. Otherwise use the formula or complete the square (see the quadratic equations guide).
3. Sketch the curve. A quadratic with a positive x² term is a U shape cutting the x-axis at the critical values.
4. Read the answer: for “> 0” take the parts of the curve above the axis, and for “< 0” take the part below it.

A “greater than” quadratic gives two separate regions (outside the critical values). A “less than” quadratic gives one region between them.

Worked example 6: greater than zero

Solve x² − 3x − 18 > 0.

x² − 3x − 18 = (x − 6)(x + 3)
Critical values: x = 6 and x = −3

The U-shaped curve is above the axis to the left of −3 and to the right of 6. The inequality is strict, so both circles are open.

x < −3 or x > 6

Never write −3 > x > 6. No number can be both less than −3 and greater than 6, so write the two parts joined with “or”.

Worked example 7: rearranging first

Solve x² + 2x ≤ 15.

x² + 2x − 15 ≤ 0
(x + 5)(x − 3) ≤ 0
Critical values: x = −5 and x = 3

“≤ 0” means on or below the axis, which is between the critical values. The boundary is included.

−5 ≤ x ≤ 3

Worked example 8: a coefficient in front of x²

Solve 2x² − x − 6 < 0.

2x² − x − 6 = (2x + 3)(x − 2)
Critical values: x = −3/2 and x = 2

The curve is below the axis between them, so −3/2 < x < 2.

Worked example 9: critical values that are not whole numbers

Solve x² − 4x − 3 ≥ 0, giving the answer in exact form.

The expression does not factorise. Solve x² − 4x − 3 = 0 with the formula:

x = (4 ± √(16 + 12)) / 2 = (4 ± √28) / 2 = 2 ± √7

The curve is on or above the axis outside the critical values:

x ≤ 2 − √7 or x ≥ 2 + √7

(2 − √7 is about −0.65 and 2 + √7 is about 4.65, so a quick sketch confirms the shape.)

Inequalities on a graph (Extension only)

A linear inequality in x and y describes a region of the plane. Its boundary is a straight line.

Drawing a region

1. Draw the boundary line, replacing the inequality sign with =.
2. Use a solid line for ≤ or ≥ (boundary included).
3. Use a dashed line for < or > (boundary not included).
4. Pick a test point that is not on the line, such as (0, 0) when the line does not pass through it. Substitute it.
5. A true result means the required region lies on the same side as the test point. A false result means it lies on the opposite side.
6. Shade the region, or shade the unwanted side if the question asks you to leave the required region unshaded. Read the question for which one.

For y > mx + c, the region is above the line. For y < mx + c, it is below.

Worked example 10: three inequalities

Show on a graph the region R where y ≥ 2, y < x + 1 and 2x + y ≤ 10.

Boundary lines:

Inequality	Line	Style	Two points
y ≥ 2	y = 2	Solid	(0, 2), (5, 2)
y < x + 1	y = x + 1	Dashed	(0, 1), (4, 5)
2x + y ≤ 10	2x + y = 10	Solid	(0, 10), (5, 0)

Test (3, 3) for y < x + 1: 3 < 4 is true, so the region is below that line. Test (0, 0) for 2x + y ≤ 10: 0 ≤ 10 is true, so the region is below that line too. For y ≥ 2 the region is on or above y = 2.

The lines cross at (1, 2), (4, 2) and (3, 4). Region R is the triangle with those corners. Its edges along y = 2 and 2x + y = 10 are included. Its edge along y = x + 1 is not.

Points with whole-number coordinates inside R: (2, 2), (3, 2), (3, 3), (4, 2). The point (1, 2) is a corner on the dashed line, so it is not in R. The point (3, 4) is also on the dashed line, so it is not in R.

Worked example 11: writing inequalities for a region

A triangular region has corners at (1, 1), (7, 1) and (3, 5), and all its edges are included. Write the three inequalities.

• Bottom edge: y = 1. The region is above it, so y ≥ 1.
• Left edge, from (1, 1) to (3, 5): gradient (5 − 1)/(3 − 1) = 2, so y = 2x − 1. A point inside, (3, 2), has 2 ≤ 2(3) − 1 = 5, so the region is below the line: y ≤ 2x − 1.
• Right edge, from (7, 1) to (3, 5): the x and y coordinates add to 8, so x + y = 8. Point (3, 2) gives 5 ≤ 8, so x + y ≤ 8.

If the left edge had been left out of the region, only that line would turn dashed and the sign would become y < 2x − 1.

Common errors

• Forgetting to reverse the sign when dividing or multiplying by a negative number.
• Using a closed circle for < or > (or an open one for ≤ or ≥).
• Writing the answer to x² > k as a single middle region. It is two outside regions.
• Giving only one root of a quadratic inequality, such as x > 6, and dropping the other region.
• Using a solid line for a strict inequality on a graph, or a dashed line for an included boundary.
• Shading the wrong side because the test point lay on the line itself.
• Including an integer that sits exactly on an open circle.

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.