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how to solve absolute value inequalities

To solve absolute value inequalities, you turn the one inequality into two related inequalities and then solve each one, being careful about signs and when to “flip” the inequality symbol.

What an absolute value inequality means

  • ∣X∣|X|∣X∣ measures the distance of XXX from 0 on the number line.
  • An inequality like ∣X∣≤a|X|\le a∣X∣≤a means “the distance from 0 is at most aaa,” so XXX must lie between −a-a−a and aaa.
  • An inequality like ∣X∣≥a|X|\ge a∣X∣≥a means “the distance from 0 is at least aaa,” so XXX is at or beyond −a-a−a and aaa, out in the “tails.”

This distance idea is the shortcut story behind all the rules.

Step-by-step: “less than” type (∣X∣≤a|X|\le a∣X∣≤a or ∣X∣<a|X|<a∣X∣<a)

These give a between (AND) answer.

General pattern

If a>0a>0a>0 and you have ∣X∣≤a|X|\le a∣X∣≤a or ∣X∣<a|X|<a∣X∣<a:

  • Replace it with a compound inequality :

−a≤X≤a-a\le X\le a−a≤X≤a

  • Then solve that compound inequality as usual.

Example 1: ∣2x−3∣≤5|2x-3|\le 5∣2x−3∣≤5

  1. Write the compound inequality:

−5≤2x−3≤5-5\le 2x-3\le 5−5≤2x−3≤5

  1. Add 3 to all three parts:

−2≤2x≤8-2\le 2x\le 8−2≤2x≤8

  1. Divide all three parts by 2:

−1≤x≤4-1\le x\le 4−1≤x≤4

So the solution is all xxx between −1-1−1 and 444, inclusive.

Step-by-step: “greater than” type (∣X∣≥a|X|\ge a∣X∣≥a or ∣X∣>a|X|>a∣X∣>a)

These give an outside (OR) answer.

General pattern

If a>0a>0a>0 and you have ∣X∣≥a|X|\ge a∣X∣≥a or ∣X∣>a|X|>a∣X∣>a:

  • Replace it with two inequalities joined by OR:

X≥aorX≤−aX\ge a\quad \text{or}\quad X\le -aX≥aorX≤−a

Example 2: ∣x+1∣>3|x+1|>3∣x+1∣>3

  1. Split into two cases:
    • x+1>3x+1>3x+1>3
    • x+1<−3x+1<-3x+1<−3
  2. Solve each:
    • x+1>3⇒x>2x+1>3\Rightarrow x>2x+1>3⇒x>2
    • x+1<−3⇒x<−4x+1<-3\Rightarrow x<-4x+1<−3⇒x<−4

So the solution is x>2x>2x>2 or x<−4x<-4x<−4.

Important first move: isolate the absolute value

Before using the patterns above, get ∣ ⋅ ∣|;\cdot;|∣⋅∣ by itself on one side.

Example 3: 3∣2x−5∣+12<393|2x-5|+12<393∣2x−5∣+12<39

  1. Subtract 12:

3∣2x−5∣<273|2x-5|<273∣2x−5∣<27

  1. Divide by 3:

∣2x−5∣<9|2x-5|<9∣2x−5∣<9

  1. Now it’s a “less than” type, so write the compound inequality:

−9<2x−5<9-9<2x-5<9−9<2x−5<9

  1. Add 5:

−4<2x<14-4<2x<14−4<2x<14

  1. Divide by 2:

−2<x<7-2<x<7−2<x<7

So the solution is all xxx strictly between −2-2−2 and 777.

Special cases: what if the right side is negative?

Because absolute value is never negative, some inequalities have no solution or are always true.

Let ∣X∣|X|∣X∣ be any absolute value.

  • If you get ∣X∣<|X|<∣X∣< (or ≤\le ≤) a negative number, there is no solution , because a distance cannot be less than a negative.
* Example: ∣2x+1∣<−4|2x+1|<-4∣2x+1∣<−4 → no solution.
  • If you get ∣X∣≥|X|\ge ∣X∣≥ (or ≥\ge ≥) a negative number, it is always true , because ∣X∣|X|∣X∣ is always ≥0\ge 0≥0.
* Example: ∣x−2∣≥−3|x-2|\ge -3∣x−2∣≥−3 → all real numbers.

Teachers often summarize this as: “Less-than with a negative → none; greater- than with a negative → all.”

Quick memory hooks (the “story” version)

Many students remember the two main patterns with a simple phrase and picture.

  • ∣X∣≤a|X|\le a∣X∣≤a: LESS than → BETWEEN
    • Solution is a segment between two points on the number line.
  • ∣X∣≥a|X|\ge a∣X∣≥a: GREATER than → OUTSIDE
    • Solution is the two outside rays going left and right.

One more way to recall it:

“Less” hugs the middle, “greater” runs to the edges.

Tiny practice set

Try these mentally or on paper using the rules above.

  1. ∣x−4∣≤6|x-4|\le 6∣x−4∣≤6
    • Rewrite as −6≤x−4≤6-6\le x-4\le 6−6≤x−4≤6 and solve.
  1. ∣3x+2∣>7|3x+2|>7∣3x+2∣>7
    • Rewrite as 3x+2>73x+2>73x+2>7 or 3x+2<−73x+2<-73x+2<−7 and solve.
  1. 5∣x+1∣−10≤05|x+1|-10\le 05∣x+1∣−10≤0
    • Isolate absolute value first: 5∣x+1∣≤10⇒∣x+1∣≤25|x+1|\le 10\Rightarrow |x+1|\le 25∣x+1∣≤10⇒∣x+1∣≤2, then use the “between” rule.

HTML table: patterns at a glance

html

<table>
  <thead>
    <tr>
      <th>Type of inequality</th>
      <th>Rewrite rule</th>
      <th>Kind of solution</th>
      <th>Example result</th>
    </tr>
  </thead>
  <tbody>
    <tr>
      <td>|X| ≤ a (a &gt; 0)</td>
      <td>-a ≤ X ≤ a</td>
      <td>Between (AND)</td>
      <td>|2x - 3| ≤ 5 → -1 ≤ x ≤ 4[web:5]</td>
    </tr>
    <tr>
      <td>|X| &lt; a (a &gt; 0)</td>
      <td>-a &lt; X &lt; a</td>
      <td>Between (AND)</td>
      <td>|x + 2| &lt; 4 → -6 &lt; x &lt; 2[web:5]</td>
    </tr>
    <tr>
      <td>|X| ≥ a (a &gt; 0)</td>
      <td>X ≥ a or X ≤ -a</td>
      <td>Outside (OR)</td>
      <td>|x + 1| ≥ 3 → x ≥ 2 or x ≤ -4[web:5][web:7]</td>
    </tr>
    <tr>
      <td>|X| &gt; a (a &gt; 0)</td>
      <td>X &gt; a or X &lt; -a</td>
      <td>Outside (OR)</td>
      <td>|x - 5| &gt; 2 → x &gt; 7 or x &lt; 3[web:5]</td>
    </tr>
    <tr>
      <td>|X| ≤ negative</td>
      <td>No solution</td>
      <td>None</td>
      <td>|x + 1| ≤ -2 → no solution[web:3][web:5]</td>
    </tr>
    <tr>
      <td>|X| ≥ negative</td>
      <td>All real numbers</td>
      <td>Always true</td>
      <td>|x - 2| ≥ -3 → all real x[web:5][web:7]</td>
    </tr>
  </tbody>
</table>

TL;DR:

  1. Isolate the absolute value.
  2. Decide if it’s a “less than” (between) or “greater than” (outside) type.
  3. Rewrite as a compound inequality (AND or OR), solve each piece, and watch out for negative right-hand sides.

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