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The Modulus Function

Q: What does the modulus of a number mean?

It is the number’s size ignoring sign — its distance from zero. So |-7| = 7 and |7| = 7. The result is always zero or positive.

Q: Why do modulus equations usually have two solutions?

Because |A| = b means A is a distance b from zero, and there are two numbers that far away: A = b and A = -b. Solving both cases (and checking) gives the two solutions.

Q: How do I solve a modulus inequality like |x - 2| < 5?

Read it as distance: “x is less than 5 from 2”, so -5 < x - 2 < 5, giving -3 < x < 7. For “greater than”, you get two outer regions instead of a band.

The modulus $|x|$ measures distance from zero. OCR uses this idea in equations and inequalities and in V-shaped graphs of the modulus of a linear function with a single modulus sign.

30 min Video by TLMaths Functions

The big picture

A modulus measures distance, so an equation usually asks for points on either side of a centre. A V-shaped graph makes those two cases visible and lets you solve equations or inequalities by locating intersections. Keep the graph and the algebra connected, and always check the domain and sign of the other side.

A-Level Maths B7-05: Introducing the Modulus FunctionCurated by Kepler, not produced by us. Watch it, then work through the notes below.

Open on YouTube

What you'll be able to do

Interpret $|x|$ as distance from zero and evaluate modulus expressions
Sketch $y=|ax+b|$ and use intersections to solve simple equations and inequalities
Solve modulus equations using cases or a sketch
Solve modulus inequalities and express the solution as a band

What the modulus means

The $\textbf{modulus}$ (or absolute value) $|x|$ is $x$ with any negative sign stripped away: $|5| = 5$ and $|-5| = 5$ . Formally it is defined in two pieces — it leaves non-negative numbers alone and flips negative ones. Best of all, read it as $\textbf{distance from zero}$ , which is always positive or zero.

That distance reading is the key to the whole topic. $|x - 3|$ is “the distance between $x$ and 3”, so $|x - 3| = 5$ means “ $x$ is 5 away from 3”, giving $x = 8$ or $x = -2$ .

|x| = \begin{cases} x & x \geq 0 \\ -x & x < 0 \end{cases}

Non-negatives stay; negatives flip sign. Result is always

\geq 0

The graphs

Start with the straight line $y=ax+b$ . Reflect the portion below the x-axis upwards; keep the rest. The result is $y=|ax+b|$ , with vertex at $(-b/a,0)$ if $a\neq0$ .

Draw the second side of an equation as another graph. The x-coordinates of intersections are solutions; for an inequality, identify the intervals where the V is above or below that other graph.

Example 1Describe how to sketch

y = |x - 2|

from

y = x - 2

1Sketch the line

y = x - 2

(crosses the

x

-axis at

x = 2

2The part below the

x

-axis (where

x < 2

) is reflected up.

3The result is a V with its vertex at

(2, 0)

AnswerA V-shape with minimum point

(2, 0)

Why this works

At the vertex the inside expression changes sign. That is exactly the split used in an algebraic piecewise definition.

Solving modulus equations

For $b>0$ , $|A|=b$ means $A=b$ or $A=-b$ . If $b=0$ , the only possibility is $A=0$ ; if $b<0$ , there is no solution because a modulus is non-negative. Split into the appropriate cases, then check every candidate back in the original equation.

For an equation like $|f(x)| = |g(x)|$ , a clean trick is to square both sides (since both are non-negative), turning it into an ordinary equation with no modulus at all.

Example 1Solve

|2x - 1| = 5

1Case 1:

2x - 1 = 5 \Rightarrow 2x = 6 \Rightarrow x = 3

2Case 2:

2x - 1 = -5 \Rightarrow 2x = -4 \Rightarrow x = -2

3Check both in the original:

|2(3)-1| = 5

✓ and

|2(-2)-1| = |-5| = 5

✓.

Answer

x = 3

x = -2

Tip — A sketch is often faster than casework: draw $y = |f(x)|$ and the horizontal line, and the intersections are your solutions — and it shows you how many to expect.

Modulus inequalities

The distance picture makes inequalities intuitive. $|x| < b$ means “within $b$ of zero”, i.e. $-b < x < b$ — a band. $|x| > b$ means “further than $b$ from zero”, i.e. $x < -b$ or $x > b$ — two outer regions. Apply the same idea to $|x - a|$ : “within $b$ of $a$ ”.

|x| < b \;\Leftrightarrow\; -b < x < b

A band around zero.

|x| > b \;\Leftrightarrow\; x < -b \text{ or } x > b

Two outer regions.

Example 1Solve

|x - 4| \leq 3

1Read it as “

x

is at most 3 from 4”.

2So

-3 \leq x - 4 \leq 3

3Add 4 throughout:

1 \leq x \leq 7

Answer

1 \leq x \leq 7

Think like an examiner

Check the sign of the right side before splitting $|A|=b$ into cases. For positive $b$ , solve both $A=b$ and $A=-b$ ; a zero or negative right side needs separate treatment.
Always check solutions in the original equation — casework can generate values that do not actually work.
Translate an inequality via the distance idea: $|x - a| < b$ is the band $a - b < x < a + b$ .

Common misconceptions

Solving

|2x - 1| = 5

and giving only

x = 3

There are two cases:

2x - 1 = \pm 5

, giving

x = 3

and

x = -2

Treating

y = |f(x)|

and

y = f(|x|)

as the same graph.

Modulus outside reflects below-axis parts up; modulus inside makes the graph symmetric in the

y

-axis.

Writing

|x| > 3

as a single band

-3 < x < 3

“Greater than” gives two outer regions:

x < -3

x > 3

Forgetting that

|x|

can never be negative.

An equation like

|x| = -2

has no solution — a distance cannot be negative.

Modulus essentials

|x| \geq 0

Distance is never negative.

|A| = b \Rightarrow A = \pm b

Two cases when solving.

|x-a| < b \Leftrightarrow a-b < x < a+b

Inequality as a band.

|x-a| > b \Leftrightarrow x < a-b \text{ or } x > a+b

Inequality as outer regions.

Stretch yourself

Solve $|2x-1|=x+2$ using a graph or cases.

Hint — The V has vertex at $x=1/2$ . Split the cases there and check the linear right side.

Questions students ask

What does the modulus of a number mean?

It is the number’s size ignoring sign — its distance from zero. So

|{-7}| = 7

and

|7| = 7

. The result is always zero or positive.

Why do modulus equations usually have two solutions?

Because

|A| = b

means

A

is a distance

b

from zero, and there are two numbers that far away:

A = b

and

A = -b

. Solving both cases (and checking) gives the two solutions.

How do I solve a modulus inequality like

|x - 2| < 5

Read it as distance: “

x

is less than 5 from 2”, so

-5 < x - 2 < 5

, giving

-3 < x < 7

. For “greater than”, you get two outer regions instead of a band.

Key takeaways

$|x|$ is distance from zero — always non-negative.
$|A| = b$ splits into $A = b$ and $A = -b$ ; always check solutions.
$|x - a| < b$ is a band $a-b < x < a+b$ ; “ $>$ ” gives two outer regions.
Squaring both sides clears the modulus when both sides are non-negative.

How this fits the course

Build on

Leads to

Partial Fractions

Test yourself

Answer a few questions, then review anything you missed.

Next step

Keep practising

Start with the quiz above to check the key ideas. You can also choose another topic from the practice library.

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