Integrated Math 2–3 Honors · Lesson Review · August 27, 2026

Solving Absolute-Value Inequalities

Translate an absolute-value inequality into an AND or OR compound inequality, solve it carefully, and write the solution with inequalities, a number-line graph, and interval notation.

16-minute edited lesson8 chaptersCaptions + transcript

01 · Watch

Lesson video

VTT captions SRT captions

0:0001Solving Absolute-Value Inequalities 0:0502What an Absolute-Value Inequality Represents 1:0903Why Less Than Becomes an AND Inequality 5:2204Example: Solve an Inward-Pointing Inequality 7:2405Why Greater Than Becomes an OR Inequality 11:1706Example: Solve an Outward-Pointing Inequality 12:5307Multi-Step AND Example 14:3508Negative-Coefficient OR Example

Video not loading? Watch directly on YouTube.

02 · Understand

Learning targets

Translate absolute-value inequalities into compound inequalities.
Choose the AND shortcut for inward-pointing inequalities.
Choose the OR shortcut for outward-pointing inequalities.
Write solution sets using inequalities, graphs, and interval notation.

03 · Vocabulary

Key definitions

Connect the algebraic shortcut to distance on a number line.

01Absolute-value inequality: An inequality containing an absolute-value expression, such as $|x|\le5$ .
02Distance meaning: $|x|$ is the distance from $x$ to 0, so it is never negative.
03AND form: When the inequality points inward, $|u|\le a$ becomes $-a\le u\le a$ .
04OR form: When the inequality points outward, $|u|\ge a$ becomes $u\le-a\text{ or }u\ge a$ .
05Union: The symbol $\cup$ joins two separate solution intervals.

04 · Apply

Worked examples

Isolate the absolute-value expression before applying a shortcut.

Why AND?

$|x|\le5$

The distance from 0 is no more than 5, so the solution stays between the two endpoints.

|x|\le5\iff-5\le x\le5\iff x\in[-5,5]

Inward → AND

$|3x+4|\le10$

$-10\le3x+4\le10$
Subtract 4 from all three parts: $-14\le3x\le6$ .
Divide all three parts by 3: $-\frac{14}{3}\le x\le2$ .

Why OR?

$|x|\ge4$

The distance from 0 is at least 4, so the solution lies on the two outside rays.

x\le-4\text{ or }x\ge4\qquad(-\infty,-4]\cup[4,\infty)

Outward → OR

$|2x+3|>5$

$2x+3>5\quad\text{or}\quad2x+3<-5$
$x>1\quad\text{or}\quad x<-4$
$(-\infty,-4)\cup(1,\infty)$

Multi-step AND

$|6-2x|+5\le7$

Isolate the absolute value: $|6-2x|\le2$ .
Use the AND form: $-2\le6-2x\le2$ .
Subtract 6: $-8\le-2x\le-4$ .
Divide by $-2$ and reverse both inequalities: $4\ge x\ge2$ .
Reorder the chain: $2\le x\le4$ .

Negative coefficient

$-3|x-5|\le-12$

Divide by $-3$ and reverse the inequality: $|x-5|\ge4$ .
$x-5\ge4\quad\text{or}\quad x-5\le-4$
$x\ge9\quad\text{or}\quad x\le1$
$(-\infty,1]\cup[9,\infty)$

05 · Avoid

Common mistakes

Using a shortcut too early.

First isolate the absolute-value expression.

Using AND for an outward inequality.

Outward-pointing inequalities create two rays joined by OR.

Forgetting the opposite endpoint.

The second boundary is the negative of the first.

Dividing by a negative without reversing.

Reverse every inequality symbol affected by the division.

Closing a strict endpoint.

Use an open circle and a parenthesis for $<$ or $>$ .

Combining OR intervals without union.

Use $\cup$ between the two separate intervals.

06 · Check yourself

Quick check

Choose AND or OR before opening each answer.

1Rewrite

|x|<7

without absolute-value bars.

$-7<x<7$ . The inequality points inward, so use AND.

2Rewrite

|x|\ge3

without absolute-value bars.

$x\le-3\text{ or }x\ge3$ . The inequality points outward, so use OR.

3Solve

|2x-1|\le7

$-3\le x\le4$ . Start with $-7\le2x-1\le7$ .

4Solve

|3x+2|>8

$x>2\text{ or }x<-\frac{10}{3}$ .

5Write

x\le1\text{ or }x\ge9

in interval notation.

$(-\infty,1]\cup[9,\infty)$ .

07 · Revisit

Transcript and assigned practice

Search the corrected transcript for an example, then complete the eKadence worksheet Absolute Value Inequalities.

Download corrected transcript