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How to Solve Inequalities and Show Them on a Number Line: A Step-by-Step GCSE Maths Guide

How to Solve Inequalities and Show Them on a Number Line: A Step-by-Step GCSE Maths Guide

Inequalities are a fundamental part of the GCSE Maths curriculum across all UK exam boards (AQA, Edexcel, OCR). They describe a range of values that satisfy a condition, rather than a single answer. Mastering them is crucial for topics like solving equations, coordinate geometry, and real-world problem-solving. This guide will walk you through solving linear inequalities and representing the solutions clearly on number lines and in regions.

What is an Inequality?

An inequality shows the relationship between two expressions that are not equal. Instead of an equals sign (=), we use symbols:

greater than

≤ less than or equal to

≥ greater than or equal to

For example, (x > 3) means "x is greater than 3". This could be 3.1, 5, or 100, but not 3 itself.

Step 1: Solving Linear Inequalities

Solving an inequality is very similar to solving an equation. The key rule is:

> You can add, subtract, multiply, or divide both sides of an inequality by the same number, just like an equation.

However, there is one critical exception:
> If you multiply or divide both sides by a negative number, you must reverse (flip) the inequality sign.

Let's work through an example.

Example 1: Solve (5x - 2 \leq 18).

Treat it like an equation. Add 2 to both sides to isolate the term with (x).


5x - 2 + 2 \leq 18 + 2


5x \leq 20


Divide both sides by 5. Since 5 is positive, the sign stays the same.


5x/5 \leq 20/5


x \leq 4


So the solution is all numbers less than or equal to 4.

Example 2: Solve ( -3x > 12 ).

Here, we need to divide by -3.

Divide both sides by -3.


-3x/-3 < 12/-3

Crucially, because we divided by a negative, we flip the > to a B{Multiply or divide
by a negative number?}
B -- No --> C[Proceed as normal.
Inequality sign stays the same.]
B -- Yes --> D[Flip the inequality sign!e.g. < becomes >, ≤ becomes ≥]
C --> E[Solution Found]
D --> E

Step 2: Representing Solutions on a Number Line

Once you have your solution (like (x \leq 4)), you need to show it visually. This is a common exam question.

Key conventions for number lines:

Closed circle (●) for ≤ (less than or equal to) or ≥ (greater than or equal to). This shows the number is included.

Open circle (○) for (greater than). This shows the number is not included.

Arrow pointing in the direction of all numbers that satisfy the inequality.

Let's show (x \leq 4) and (x < -4) from our examples.

flowchart LR
subgraph “Inequality: x ≤ 4”
direction LR
A1[“ ”] --- B1[“ ”] --- C1[“ ”] --- D1[“ ”] --- E1[“● 4”] --- F1[“ ”]
style E1 fill:#f9f,stroke:#333
linkStyle 0,1,2,3 stroke:red,stroke-width:4px
linkStyle 4 stroke:none
end

subgraph “Inequality: x < -4”
direction LR
A2[“ ”] --- B2[“○ -4”] --- C2[“ ”] --- D2[“ ”] --- E2[“ ”] --- F2[“ ”]
style B2 fill:#f9f,stroke:#333
linkStyle 5 stroke:none
linkStyle 6,7,8,9 stroke:red,stroke-width:4px
end

For (x \leq 4), we place a closed circle on 4 and draw an arrow pointing left, towards the smaller numbers.

For (x < -4), we place an open circle on -4 and draw an arrow pointing left.

Step 3: Shading Regions on Graphs

For inequalities with two variables (like (y > 2x + 1)), the solution is a set of coordinate points that form a region on a graph.

How to shade the region for (y > 2x + 1):

Treat the inequality as an equation ((y = 2x + 1)) and plot the line. This line is the boundary of the region.

Choose the line style:

Use a solid line for ≤ or ≥ (points on the line are included).

Use a dashed line for (points on the line are not included).

Since we have (>), we draw a dashed line for (y = 2x + 1).

Choose which side to shade. Pick a test point that is not on the line. The origin ((0,0)) is often easiest.

Substitute ((0,0)) into the inequality: (0 > 2(0) +

1) → (0 > 1). This is false.

Since ((0,0)) makes the inequality false, we shade the region on the opposite side of the line from ((0,0)).

Commo

Read now…
Exam question of the day
The diagram shows quadrilateral OACB.
Diagram for the exam question
M is the midpoint of OA.
N is the point on BC such that BN:NC = 4 :5
OA = a
OB = b
AC = kb where k is a positive integer.
(a) Express MN in terms of k, a and b.
Give your answer in its simplest form.
(b) Is MN parallel to OB?
Give a reason for your answer.
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