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Learn · Mathematics · Grade 9 · Algebra

Absolute value: equations and inequalities

|x| as a distance from zero, |x − a| as a distance from a, the two cases of |ax + b| = c, and why |x| < c is a strip and |x| > c is two rays.

High school Lesson 4 of 31 Interactive figure Practice

First, the rules

Absolute value is a distance

|x| is how far x is from 0, never negative. |x − a| is how far x is from a. Read every equation and inequality below as a sentence about distance and it becomes obvious.

Definition
|x| = x if x ≥ 0, |x| = −x if x < 0
|−7| = 7, |7| = 7, |0| = 0.
Properties
|ab| = |a| · |b|, |ab| = |a||b|, |a| = |−a|, |a|² = a²
★ Equation |ax + b| = c
Example 1
c < 0: no solution. c = 0: ax + b = 0. c > 0: two cases, ax + b = c or ax + b = −c.
★ |x − a| < c — "within c of a"
Example 2
a − c < x < a + c
A strip between two points.
★ |x − a| > c — "farther than c from a"
Example 2
x < a − c or x > a + c
Two rays, pointing away from a. The word is OR, never AND.
Equation with |…| on both sides
Example 3
|u| = |v| ⇔ u = v or u = −v

See it

Distance from a

Slide the centre a and the distance c. "Less than" shades the strip, "greater than" shades the two rays, "equal" marks the two endpoints.

Step by step

Worked examples

1

Two cases

equation

Solve |2x − 5| = 7.

Ask yourself: What is 7 units away from 0? Both 7 and −7. So what can 2x − 5 be?

Show the solution
  1. Case 1: 2x − 5 = 7 ⇒ 2x = 12 ⇒ x = 6.
  2. Case 2: 2x − 5 = −7 ⇒ 2x = −2 ⇒ x = −1.
  3. Check both: |12 − 5| = 7 ✓, |−2 − 5| = 7 ✓.

Answer x ∈ {−1, 6}

2

Strip and rays

inequalities

Solve (a) |x − 3| < 4 and (b) |x + 1| ≥ 2.

Ask yourself: Say each one in words: (a) "x is less than 4 away from 3"; (b) "x is at least 2 away from −1".

Show the solution
  1. (a) Within 4 of 3: 3 − 4 < x < 3 + 4, i.e. −1 < x < 7.
  2. (b) |x + 1| = |x − (−1)|: at least 2 away from −1, in either direction: x ≤ −3 or x ≥ 1.
  3. On the number line (a) is one strip with open ends, (b) is two rays with filled ends.

Answer (a) −1 < x < 7 (b) x ≤ −3 or x ≥ 1

3

Absolute values on both sides

|u| = |v|

Solve |3x − 1| = |x + 5|.

Ask yourself: Two numbers have the same absolute value when they are equal or opposite.

Show the solution
  1. Equal: 3x − 1 = x + 5 ⇒ 2x = 6 ⇒ x = 3.
  2. Opposite: 3x − 1 = −(x + 5) ⇒ 4x = −4 ⇒ x = −1.
  3. Check: x = 3 gives |8| = |8| ✓; x = −1 gives |−4| = |4| ✓.

Answer x ∈ {−1, 3}

Practice: absolute value

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Equations with two cases, strips and rays — choose the solution set. Your best score is kept in this browser only.

About this lesson

Absolute value explained as distance: |x| = x or −x, |x − a| as the distance between x and a, properties of absolute value, solving |ax + b| = c by two cases, inequalities |ax + b| < c (between) and |ax + b| > c (outside), and reading the solution on a number line you can move. Worked examples and practice.

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