Learn9MA0Paper 1 · Pure Mathematics 1Algebra and functionsModulus graphs and equations

Algebra and functions · Subtopic 9 of 16

Modulus Graphs and Equations

What $|x|$ means, how to sketch $y=|ax+b|$ by reflecting the negative part, and two reliable ways to solve modulus equations: squaring both sides, or splitting into cases and checking.

Cambridge 9709 Paper 2Cambridge 9709 Paper 3Edexcel YMA01 WMA13Edexcel 9MA0 Paper 1Edexcel 9MA0 Paper 2AQA 7357 Paper 1≈ 40 minRevision 6 · updated 2026-10-27

Section 1 of 14

Learning objectives

Understand $|x|$ as distance from $0$ , and $|a-b|$ as the distance between $a$ and $b$ .
Sketch $y=|ax+b|$ , marking the vertex and intercepts.
Solve $|ax+b|=k$ and $|ax+b|=|cx+d|$ by squaring or by cases.
Solve $|ax+b|=cx+d$ , checking for and rejecting invalid solutions.

Pause and think

On a number line, which numbers are exactly $3$ units away from $5$ ?

Section 2 of 14

Core idea

Start hereHow far, not which way

A lift is at floor $2$ . It moves to floor $-3$ (car park) or floor $7$ . How far did it travel? $5$ floors either way, regardless of direction. Distance ignores sign. The modulus $|x|$ is the mathematical "distance from zero", and it creates graphs with sharp corners and equations with two answers.

The big idea

$|x|$ is the size of $x$ without its sign: $|x|=x$ if $x\ge0$ , $|x|=-x$ if $x<0$ . $|x-a|$ is the distance between $x$ and $a$ . $|f(x)|=k$ splits into $f(x)=k$ or $f(x)=-k$ ; $y=|f(x)|$ reflects the negative parts of $y=f(x)$ upward.

1. Definition

|x|=\begin{cases}x,&x\ge0\\-x,&x<0\end{cases}

$-x$ when $x<0$: minus a negative is positive: $|-5|=-(-5)=5$ .

$|5|=5$ , $|-5|=5$ , $|0|=0$ .
$|x-a|$ = distance between $x$ and $a$ on the number line.
$|ab|=|a||b|$ but $|a+b|\le|a|+|b|$ (not always equal).

2. Graphs

Making y = |2x − 3| from y = 2x − 3

Any part of the line below the

x

-axis is reflected upwards, giving a V shape. The point of the V is where the inside,

2x-3

, equals zero.

Reflecting the negative part of a curve

flip progress = 0.00

The part of

y=x^2-4

between

-2

and

2

(below the axis) flips over the axis to make

y=|x^2-4|

To draw $y=|f(x)|$ : draw $y=f(x)$ , then reflect every part below the $x$ -axis in the axis. The vertex of $y=|2x-3|$ is where the inside is zero: $x=\frac32$ .

Edge case $|f(x)|$ versus $f(|x|)$

y=f(|x|)

is different: take the part for

x\ge0

and reflect it in the

y

-axis.

|x|^2-4|x|

is symmetric about the

y

-axis;

|x^2-4x|

is not.

3. Solving equations

Worked example $|2x-3|=7$

Solve.

Goal: All $x$ .
Strategy: The inside is $7$ or $-7$ .

$2x-3=7\Rightarrow x=5$
Positive case.
$2x-3=-7\Rightarrow x=-2$
Negative case.

Answer

x=5

x=-2

✓ Check

|7|=7

✓;

|-7|=7

✓.

Worked example $|x-1|=|2x+3|$ by squaring

Solve.

Goal: All $x$ .
Strategy: Both sides are $\ge0$ , so squaring loses nothing and gains nothing; $|A|^2=A^2$ .

$(x-1)^2=(2x+3)^2$
Square both sides.
$x^2-2x+1=4x^2+12x+9$
Expand.
$0=3x^2+14x+8=(3x+2)(x+4)$
Collect; factorise.

Answer

x=-\frac23

x=-4

✓ Check

x=-4

|-5|=5

and

|-5|=5

✓.

Worked example $|x-4|=2x+1$ : checking is essential

Solve.

Goal: All valid $x$ .
Strategy: Try both cases; reject any where the right side would be negative.

$x-4=2x+1\Rightarrow x=-5$
Case 1.
Check: RHS $=2(-5)+1=-9<0$ , but a modulus can't be negative. Reject.
$-(x-4)=2x+1\Rightarrow3=3x\Rightarrow x=1$
Case 2.
Check: $|1-4|=3$ and $2+1=3$ ✓.

Answer

x=1

only.

✓ CheckA sketch shows the line

y=2x+1

crossing the V once ✓.

Common misconceptionNot checking When the other side isn't a modulus, one case can give a false root. Always substitute back, or sketch first.

Pause and think

Solve $|3x+1|=5$ .

Summary

$|x|$ = distance from $0$ .
$|f|=k$ : two cases.
$|f|=|g|$ : square.
$|f|=g$ : check solutions.

Modulus: Absolute value.

HistoryTwo straight lines. The notation

|x|

was introduced by Karl Weierstrass in 1841. Before that, mathematicians wrote "the numerical value of

x

". Weierstrass also made calculus rigorous, and his vertical bars are now used for size everywhere: the modulus of a complex number, the magnitude of a vector, the determinant of a matrix.

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