Modulus & equations — notes, question types & practice · Functions, Graphs & Logarithms

01

Modulus means distance

$|x|$ is the distance of $x$ from $0$ . Distance is never negative, so $|-7| = 7$ and $|7| = 7$ .

$|x - 3|$ is the distance of $x$ from $3$ . The point $x = 5$ sits at distance $2$ , and so does $x = 1$ : both give $|x - 3| = 2$ .

Distance thinking solves modulus questions faster than algebra. Ask: which points sit at this distance from this number?

Rule: $|x - a| = d$ means $x$ sits at distance $d$ from $a$ — the two points $a + d$ and $a - d$ .

02

Equations: split into two cases

The inside of a modulus is either positive or negative. Solve both ways.

Solve $|2x - 3| = 5$ :

Case one: $2x - 3 = 5$ , so $x = 4$ .
Case two: $2x - 3 = -5$ , so $x = -1$ .
Both are valid: $x = 4$ or $x = -1$ .

Or use distance: $|x - 2| = 5$ asks for points at distance $5$ from $2$ , namely $7$ and $-3$ .

$|x| = -4$ has no solution — no distance is negative.

03

Inequalities: inside or outside

$|x - a| < d$ keeps $x$ within distance $d$ of $a$ — one band.

$|x - 3| < 4$ gives $-1 < x < 7$ .

$|x - a| > d$ pushes $x$ outside the band — two arms.

$|2x - 1| \ge 5$ : the quantity $2x - 1$ sits at least $5$ from zero, so $2x - 1 \ge 5$ or $2x - 1 \le -5$ . That gives $x \ge 3$ or $x \le -2$ .

Watch: "<" gives one interval, ">" gives two arms. Mixing them up is the most common modulus error in the paper.

04

Sum of two distances

$|x - a| + |x - b|$ adds the distances from $x$ to two fixed pins $a$ and $b$ .

Walk between the pins: each step moves you closer to one pin and further from the other by the same amount, so the total stays fixed inside.

\text{minimum of } |x-a| + |x-b| = |a - b|

Worked: the minimum of $|x - 4| + |x + 2|$ is the gap between the pins $4$ and $-2$ , which is $6$ . Every $x$ from $-2$ to $4$ achieves it.

Outside the pins the total grows: at $x = 10$ it is already $18$ .

Tip: For the minimum of $|x - a| + |x - b|$ , just subtract the pins. Any $x$ between them achieves it.

05

The validity check

When a modulus equals an expression containing $x$ , a case solution can be fake. Test each root in the original equation.

Solve $|x - 3| = 2x - 5$ :

Case $x \ge 3$ : $x - 3 = 2x - 5$ gives $x = 2$ . But $2 < 3$ — the root breaks its own case. Reject.
Case $x < 3$ : $3 - x = 2x - 5$ gives $3x = 8$ , so $x = \dfrac{8}{3}$ .
Check: $\left|\dfrac{8}{3} - 3\right| = \dfrac{1}{3}$ and $2 \cdot \dfrac{8}{3} - 5 = \dfrac{1}{3}$ . Valid.

Answer: $x = \dfrac{8}{3}$ alone. The right side $2x - 5$ must be non-negative for any solution, and $x = 2$ made it negative.

06

Exponential equations

Match the bases, then equate the powers.

$2^x = 64$ : since $64 = 2^6$ , $x = 6$ .

$3^{x+2} = 27$ : since $27 = 3^3$ , $x + 2 = 3$ and $x = 1$ .

Products on one side add powers. $2^x \cdot 2^{x+1} = 32$ becomes $2^{2x+1} = 2^5$ , so $2x + 1 = 5$ and $x = 2$ .

When no power matches, take logs: $3^x = 20$ gives $x = \dfrac{\log 20}{\log 3}$ .

Note: Every exponential equation is a log equation with the base matched first. Logs step in only when the bases refuse to match.

07

Question types you will see

Each type: how to recognise it, the method step by step, and one question to try.

Type 1very common5 practice Q

Solve |expression| = number

How to spot it:

An equation like $|2x - 3| = 5$ or $|x - 2| = 5$ with two solutions expected.

|x - a| = d \Rightarrow x = a \pm d

Method

Write the two cases: inside $= +d$ and $= -d$ .
Solve each linear equation.
Or use distance: $a + d$ and $a - d$ .
List both answers.

Why it works:

The inside of a modulus can be positive or negative, so both sign choices are candidates.

Try this

Solve $|2x - 3| = 5$ .

Show solution

$2x - 3 = 5$ gives $x = 4$ .
$2x - 3 = -5$ gives $x = -1$ .

Answer

x = 4 or x = -1

Type 2common2 practice Q

Modulus inequality: inside the band

How to spot it:

$|x - a| < d$ or $\le d$ — one interval is the answer.

|x - a| < d \Rightarrow a - d < x < a + d

Method

Identify the centre $a$ and radius $d$ .
Subtract: $a - d$ . Add: $a + d$ .
Write the interval between them.
Keep strict or loose signs as given.

Why it works:

Staying within distance $d$ of $a$ means living inside the band of radius $d$ .

Try this

Solve $|x - 3| < 4$ .

Show solution

Centre $3$ , radius $4$ .
$-1 < x < 7$ .

Answer

-1 < x < 7

Type 3common2 practice Q

Modulus inequality: outside the band

How to spot it:

$|x - a| > d$ or $\ge d$ — two separate arms are the answer.

|x - a| > d \Rightarrow x < a - d \text{ or } x > a + d

Method

Identify centre $a$ and radius $d$ .
Split into the two cases $\ge d$ and $\le -d$ inside.
Solve each inequality.
Join the two arms with 'or'.

Why it works:

Being further than $d$ from $a$ leaves only the two outer regions of the line.

Try this

Solve $|x + 2| > 6$ .

Show solution

$x + 2 > 6$ gives $x > 4$ .
$x + 2 < -6$ gives $x < -8$ .

Answer

x < -8 or x > 4

Type 4common2 practice Q

Minimum of |x−a| + |x−b|

How to spot it:

'The minimum value of $|x - a| + |x - b|$ ' with two fixed numbers $a$ and $b$ .

\min\big(|x-a| + |x-b|\big) = |a - b|

Method

Spot the two pins $a$ and $b$ .
Take their gap $|a - b|$ .
State that any $x$ between them achieves it.
Check one inside value if unsure.

Why it works:

Between the pins, one distance grows exactly as the other shrinks, so the sum cannot drop below the gap.

Try this

Find the minimum of $|x - 4| + |x + 2|$ .

Show solution

Pins at $4$ and $-2$ .
Gap $= 4 - (-2) = 6$ .

Answer

6

Type 5occasional2 practice Q

Equations needing a validity check

How to spot it:

A modulus equals an expression with $x$ , like $|x - 3| = 2x - 5$ . One case may fail.

Method

Split into the two sign cases.
Solve each case.
Check each root in the original equation.
Reject roots that make the non-modulus side negative or break the case.

Why it works:

A modulus is non-negative, so the other side must be too — some algebraic roots break that.

Try this

Solve $|x - 3| = 2x - 5$ .

Show solution

Case $x \ge 3$ : $x = 2$ — rejected, breaks the case.
Case $x < 3$ : $x = \dfrac{8}{3}$ .
Check: both sides equal $\dfrac{1}{3}$ .

Answer

x = 8/3

Type 6very common3 practice Q

Exponential equation by matching bases

How to spot it:

Powers like $2^x = 64$ or $3^{x+2} = 27$ — both sides can share one base.

a^{f(x)} = a^{g(x)} \Rightarrow f(x) = g(x)

Method

Write both sides as powers of the same base.
Equate the exponents.
Solve the small equation.
Products become sums of exponents.

Why it works:

Equal powers of the same base force equal exponents.

Try this

Solve $2^x \cdot 2^{x+1} = 32$ .

Show solution

Left side: $2^{2x+1}$ ; right side: $2^5$ .
$2x + 1 = 5$ .
$x = 2$ .

Answer

x = 2

08

Formula sheet

Modulus equation

|x - a| = d \Rightarrow x = a \pm d

Two points at distance $d$ from $a$.

Inside the band

|x| < a \Rightarrow -a < x < a

One interval; needs $a > 0$.

Outside the band

|x| > a \Rightarrow x < -a \text{ or } x > a

Two arms; the sign '>' splits.

Distance sum

\min\big(|x-a| + |x-b|\big) = |a - b|

Every $x$ between $a$ and $b$ achieves it.

Exponential match

a^{f(x)} = a^{g(x)} \Rightarrow f(x) = g(x)

Write both sides with the same base first.

09

Shortcuts that save time

⚡ Think distance, not cases

$|x - 3| = 2$ asks for points at distance $2$ from $3$ . Walk $2$ left and $2$ right: $x = 1$ and $x = 5$ . No case work.

Example

Solve $|x - 2| = 5$ .

Show solution

Points at distance $5$ from $2$ .
$2 + 5 = 7$ , $2 - 5 = -3$ .

Answer

x = 7 or x = -3

⚡ Band or arms — one glance

For $|x - a| < d$ the answer sits between $a - d$ and $a + d$ . For $> d$ it sits outside those two numbers.

Example

Solve $|x - 3| < 4$ .

Show solution

Band around $3$ at radius $4$ .
$3 - 4 = -1$ , $3 + 4 = 7$ .

Answer

-1 < x < 7

⚡ Subtract the pins for the minimum

The least value of $|x - a| + |x - b|$ is the gap between the pins. One subtraction, done.

Example

Find the minimum of $|x - 1| + |x - 7|$ .

Show solution

Pins at $1$ and $7$ .
Gap $= 7 - 1 = 6$ .

Answer

6

10

Mistakes to avoid

Where most students lose marks on this subtopic.

Mistake 01

Solving $|x| = -4$ and writing $x = \pm 4$ — a modulus is never negative, so there is no solution.

Mistake 02

Splitting $|x - 2| < 5$ as $x < 7$ only — the other side gives $x > -3$ too.

Mistake 03

Writing $-a < x < a$ for $|x| > a$ — that is the band for '<'; the sign '>' gives two arms outside.

Mistake 04

Accepting a case root without checking it in the original equation — fake roots appear when the right side contains $x$ .

Mistake 05

Putting the minimum of $|x-a| + |x-b|$ only at $x = a$ — every point between the pins gives the same minimum.

Mistake 06

Solving $2^{x+1} = 8$ as $x + 1 = 8$ — first write $8 = 2^3$ , then equate powers: $x + 1 = 3$ .

11

Quick revision

Read this the night before the exam.

$|x - a| = d$ : two answers, $a + d$ and $a - d$ .
$|x - a| < d$ : one band; $|x - a| > d$ : two arms.
$\min(|x-a| + |x-b|) = |a - b|$ , true for all $x$ between the pins.
Check every root when a modulus equals an expression in $x$ .
$a^{f(x)} = a^{g(x)}$ gives $f(x) = g(x)$ after matching bases.
$|x| =$ negative has no solution.

12

Practice: 16 questions

Sets of 10, mixed across the question types above. Every answer has a step-by-step explanation.

Topic test · 10 questions

Suggested time 6 min · wrong answers go to your mistake notebook automatically.