How do I solve an equation with an absolute value?

For c > 0 the equation |W(x)| = c splits into two branches, W(x) = c and W(x) = −c, each solved separately. For c = 0 a single equation W(x) = 0 is left, and for c < 0 there are no solutions at all, because a modulus is never negative.

How many solutions does |x − a| = r have?

For r > 0 exactly two: a − r and a + r. For r = 0 exactly one, namely a itself. For r < 0 none. It follows straight from the distance reading: there are always two points a given positive distance from a — one on the left, one on the right.

What is the difference between |x − a| r?

The first describes numbers closer than r to a, which is one interval (a − r, a + r). The second describes numbers further than r away, which is everything outside it — the union (−∞, a − r) ∪ (a + r, +∞). Writing the second case as a single interval is the most common mistake in the topic.

How do I write an interval as one condition with a modulus?

The centre of [p, q] is a = (p + q)/2 and its radius is r = (q − p)/2. The interval is then |x − a| ≤ r. For [4, 10] we get a = 7 and r = 3, so the condition reads |x − 7| ≤ 3.

When does an inequality with a modulus have no solutions?

When it demands of a modulus something a modulus cannot do. |x + 1| −2 is satisfied by every real number. Before splitting a modulus into cases it is worth checking the sign on the right-hand side.

Advanced level

Absolute value equations and inequalities

Q: How many solutions does |x − a| = r have?

For r > 0 exactly two: a − r and a + r. For r = 0 exactly one, namely a itself. For r < 0 none. It follows straight from the distance reading: there are always two points a given positive distance from a — one on the left, one on the right.

Q: What is the difference between |x − a| r?

The first describes numbers closer than r to a, which is one interval (a − r, a + r). The second describes numbers further than r away, which is everything outside it — the union (−∞, a − r) ∪ (a + r, +∞). Writing the second case as a single interval is the most common mistake in the topic.

Q: How do I write an interval as one condition with a modulus?

The centre of [p, q] is a = (p + q)/2 and its radius is r = (q − p)/2. The interval is then |x − a| ≤ r. For [4, 10] we get a = 7 and r = 3, so the condition reads |x − 7| ≤ 3.

Q: When does an inequality with a modulus have no solutions?

When it demands of a modulus something a modulus cannot do. |x + 1| −2 is satisfied by every real number. Before splitting a modulus into cases it is worth checking the sign on the right-hand side.

The modulus measures a distance — and that one sentence solves the whole lesson. The equation |x − a| = r asks for the points r away from a, a strict inequality gives an interval around a, and the other direction gives a union of two. Plus reading the centre and radius back, and the cases with no solution at all.

Before you start

This topic builds on earlier ideas. Before you start, it's worth working through the lessons below — they'll make everything click:

Where this is used

Real situations where you count exactly the way this lesson teaches:

A parcel and its weight limit
A carrier accepts parcels of 5 kg with a 200 g tolerance, which is the single condition |m − 5| ≤ 0.2 and means the range from 4.8 to 5.2 kg. A parcel weighing 5.25 kg falls into the next price band — over 50 grams nobody can see by eye.
Running and training pace
A plan calls for 5:30 per kilometre with a 15-second tolerance, that is |t − 330| ≤ 15 in seconds per kilometre — from 5:15 to 5:45. Over 10 km that spreads the finish time from 52:30 to 57:30, five minutes across which a session still counts as "about the same".
Convert it here: Pace
A lathe and a dimensional tolerance
A drawing specifies a shaft at 20 mm with a tolerance of ±0.05 mm, that is |d − 20| ≤ 0.05. A part measuring 20.03 mm passes inspection and one at 20.08 mm is scrapped — the whole decision fits inside one inequality with a modulus.
Convert it here: Length
Quality control and the three-sigma rule
A machine packs coffee at 500 g with a standard deviation of 4 g. The alarm is set on the condition |m − 500| > 12, that is outside the range [488, 512] grams. On a machine working correctly such a pack turns up less than once in 300, so every one of them is a reason to stop the line.

All formulas

Equation with a modulus
$|x| = c \iff x = c \ \text{or} \ x = -c, \quad c > 0$
for c = 0 one solution, for c < 0 none
Distance equal to r
$|x - a| = r \iff x = a - r \ \text{or} \ x = a + r$
two points r away from a
Closer than r
$|x - a| < r \iff x \in (a - r, a + r)$
one interval around the centre a
Further than r
$|x - a| > r \iff x \in (-\infty, a - r) \cup (a + r, +\infty)$
a union of two intervals, never one
Modulus of a linear expression
$|ax + b| = c \iff ax + b = c \ \text{or} \ ax + b = -c$
two branches, each solved on its own
Centre and radius of an interval
$a = \frac{p + q}{2}, \qquad r = \frac{q - p}{2}$
writing the interval [p, q] as the single condition |x − a| ≤ r

Absolute value measures a distance on the number line: $|x - a|$ is how far $x$ is from $a$ . This whole lesson unfolds from that one sentence — an equation with a modulus asks for the points at a given distance, and an inequality asks for the points closer or further than a given distance.

You can also split the modulus into cases from the definition, and sometimes you must. But in ordinary school problems the distance reading is faster and, more importantly, it does not let you lose the second solution.

The equation |x| = c

|x| = c \iff x = c \ \text{or} \ x = -c, \qquad c > 0

A modulus does not distinguish signs, so the equation has two solutions symmetric about zero. The other two cases follow from the fact that a modulus is never negative:

$|x| = 0$ has one solution: $x = 0$ ;
$|x| = -3$ has none, because the left side is non-negative and the right side is negative.

Shifting the centre changes nothing. The equation $|x - a| = r$ asks for the numbers $r$ away from $a$ , and there are two of them — one on each side:

|x - a| = r \iff x = a - r \ \text{or} \ x = a + r

The equation |x + 1| = 4. Since x + 1 is x − (−1), the centre is −1, and four steps each way give −5 and 3.

It is worth reading $|x + 1|$ straight away as $|x - (-1)|$ — otherwise it is easy to take $1$ for the centre instead of $-1$ .

The modulus of a linear expression

When something more than $x$ sits inside the modulus, the equation splits into two branches, each solved on its own:

|ax + b| = c \iff ax + b = c \ \text{or} \ ax + b = -c

Solve |2x − 6| = 8.

Inequalities: one band or two tails

This is where half the answer is most easily lost. The direction of the inequality decides the shape of the solution set:

|x - a| < r \iff x \in (a - r, a + r)

|x - a| > r \iff x \in (-\infty, a - r) \cup (a + r, +\infty)

The first says "closer than $r$ to $a$ " and gives one interval around the centre. The second says "further than $r$ from $a$ " and gives a union of two intervals — everything outside that band. A non-strict sign ( $\le$ , $\ge$ ) only closes the ends; it does not change the shape.

The band on the axis is the solution of |x − 2| ≤ 3. The two bands above it solve |x − 2| > 3 — together they cover the whole axis, because they are complementary sets.

The two sets together cover the whole axis and do not overlap. That is the simplest way to check an answer: if the two solution sets do not complement each other, something went wrong.

Solve |3x + 6| ≥ 9.

Centre and radius: the other direction

The same pair of formulas read backwards turns an interval into a single condition with a modulus. The centre of $[p, q]$ is its midpoint and the radius is half its length:

a = \frac{p + q}{2}, \qquad r = \frac{q - p}{2}

For the interval $[4, 10]$ that gives $a = 7$ and $r = 3$ , so the condition $4 \le x \le 10$ becomes $|x - 7| \le 3$ . This is how every tolerance is written: the nominal value is the centre, the tolerance is the radius.

Cases with no solutions

Before splitting a modulus, it is worth looking at the right-hand side:

Inequality	Solution set	Why
$\lvert W(x) \rvert < -2$	$\varnothing$	a modulus is never negative
$\lvert W(x) \rvert > -2$	$\mathbb{R}$	a modulus always exceeds a negative number
$\lvert W(x) \rvert \le 0$	the solutions of $W(x) = 0$	a modulus is zero only at zero
$\lvert W(x) \rvert \ge 0$	$\mathbb{R}$	the condition always holds

Four rows that need no computation — only recognition.

Practice

Work through a set of exercises — they get harder as you go. At the end you'll see your score and the mistakes worth reviewing.

Exercise 1 of 8Score: 0

|x − 2| = 2

Common mistakes

One solution instead of two — $|x| = 5$ has the solutions $5$ and $-5$ . A modulus does not distinguish signs, so keeping only the positive one is not allowed.
Writing $\lvert x - a \rvert > r$ as a single interval — the answer is always a union of two intervals, one on each side.
The wrong sign of the centre — in $|x + 1|$ the centre is $-1$ , not $1$ . Rewriting it as $|x - (-1)|$ helps.
Not reversing the second branch — from $|W| \ge c$ the second branch is $W \le -c$ , with the inequality reversed.
Solving an equation with a negative right-hand side — $|x - 3| = -2$ has no solutions, and there is nothing to split into cases.
Dropping the coefficient when factoring it out — $|3x + 6|$ is $3|x + 2|$ , so the inequality $\ge 9$ becomes $|x + 2| \ge 3$ , not $\ge 9$ .

Formula card

Topic: Absolute value equations and inequalities

Equation with a modulus
$|x| = c \iff x = c \ \text{or} \ x = -c, \quad c > 0$
for c = 0 one solution, for c < 0 none
Distance equal to r
$|x - a| = r \iff x = a - r \ \text{or} \ x = a + r$
two points r away from a
Closer than r
$|x - a| < r \iff x \in (a - r, a + r)$
one interval around the centre a
Further than r
$|x - a| > r \iff x \in (-\infty, a - r) \cup (a + r, +\infty)$
a union of two intervals, never one
Modulus of a linear expression
$|ax + b| = c \iff ax + b = c \ \text{or} \ ax + b = -c$
two branches, each solved on its own
Centre and radius of an interval
$a = \frac{p + q}{2}, \qquad r = \frac{q - p}{2}$
writing the interval [p, q] as the single condition |x − a| ≤ r

The equation |x + 1| = 4 asks for the numbers 4 away from −1. The answer is two points: −5 and 3.

The inequality |x − 2| ≤ 3 is a band around 2, and |x − 2| > 3 is everything outside it — two intervals, not one.

Frequently asked questions

A balance scale in perfect equilibrium rendered as a geometric form, with a straight line crossing an axis like an orbital path, against a cosmic nebula in violet and magenta

Algebra

What x really means — unknown, variable and parameter

The letter x does not always mean the same thing. Sometimes it is a number you are hunting for, sometimes a quantity that varies continuously, sometimes a blank waiting for any number at all. Here is where algebraic notation came from, why it took two thousand years to arrive, what exactly separates an unknown from a variable and a parameter, and why an inequality flips when you multiply by a negative number.

11 min