Learn7357Paper 1 · PureAlgebra and functionsModulus inequalities

Algebra and functions · Subtopic 4 of 9

Modulus Inequalities

Solving inequalities with moduli: the "between" and "outside" patterns, squaring when both sides are moduli, and using a sketch when a modulus is compared with a line.

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

Rewrite $|x-a|<k$ as $a-k<x<a+k$ , and $|x-a|>k$ as $x<a-k$ or $x>a+k$ .
Solve $|A|<|B|$ by squaring and solving a quadratic inequality.
Solve $|A|<B$ (with $B$ a line) using the critical values and a sketch.
Write solutions in correct inequality form.

Pause and think

Which numbers are less than $3$ units from $1$ on the number line? Which are more than $3$ units away?

Section 2 of 14

Core idea

Start hereWithin tolerance

A machine cuts rods meant to be $50$ mm long, accurate to within $0.2$ mm. Acceptable lengths $L$ satisfy $|L-50|\le0.2$ : "the distance from $50$ is at most $0.2$ ", i.e. $49.8\le L\le50.2$ . Modulus inequalities are statements about distance.

The big idea

$|x-a|<b\iff a-b<x<a+b$ (within $b$ of $a$ ); $|x-a|>b\iff x<a-b$ or $x>a+b$ (further than $b$ ). For $|f(x)|<|g(x)|$ , square both sides.

1. Distance on the number line

|x − 2| < 3 on a number line

Everything within

3

2

: the open interval

-1<x<5

Worked example $|2x-5|\le7$

Solve.

Goal: An interval.
Strategy: Unpack to a double inequality.

$-7\le2x-5\le7$
Within $7$ of $0$ .
$-2\le2x\le12$
Add $5$ to all three parts.
$-1\le x\le6$
Halve.

Answer

-1\le x\le6

✓ Check

x=6

|7|=7\le7

✓;

x=7

9>7

✓ excluded.

Worked example $|x+3|>4$

Solve.

Goal: Two pieces.
Strategy: More than $4$ away from $-3$ .

$x+3>4\ \text{or}\ x+3<-4$
Outside.
$x>1\ \text{or}\ x<-7$
Solve each.

Answer

x<-7

x>1

✓ Check

x=0

3>4

false ✓ (correctly excluded).

2. Two moduli: square

Where is |x − 1| below |2x + 3|?

The graph of

|x-1|

is below

|2x+3|

to the left of

-4

and to the right of

-\tfrac23

, matching the algebra.

Worked example $|x-1|<|2x+3|$

Solve.

Goal: Set of $x$ .
Strategy: Both sides $\ge0$ , so $|A|<|B|\iff A^2<B^2$ .

$x^2-2x+1<4x^2+12x+9$
Square and expand.
$0<3x^2+14x+8=(3x+2)(x+4)$
Rearrange; factorise.
$x<-4\ \text{or}\ x>-\tfrac23$
∪ shape positive outside the roots.

Answer

x<-4

x>-\frac23

✓ Check

x=0

1<3

✓;

x=-2

3<1

false ✓.

3. Modulus against a line: use the graph

Worked example $|x-2|<x$

Solve.

Goal: Set of $x$ .
Strategy: Find where they're equal, then use a sketch.

$-(x-2)=x\Rightarrow x=1$
Left branch meets $y=x$ .
$x-2=x\Rightarrow-2=0$
Right branch: never meets (parallel).
Sketch: $y=x$ is above the V to the right of $x=1$ .

Answer

x>1

✓ Check

x=3

1<3

✓;

x=0

2<0

false ✓.

Common misconceptionSquaring when one side might be negative Only square both sides when both are non-negative.

|x-2|<x

: the right side can be negative, so use the graph.

Pause and think

Solve $|x|<|x-6|$ .

Summary

$<$ : between; $>$ : outside.
Two moduli: square.
Otherwise: critical values + sketch.

Critical value: Where the two sides are equal.

HistoryInequality symbols. The symbols

<

and

>

first appeared in a book by Thomas Harriot published in 1631, ten years after his death. Harriot was an astronomer who drew the Moon through a telescope months before Galileo, but published very little in his lifetime.

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