Absolute Value Inequalities Notes Absolute Value Inequalities - Notes
Name:
Determining the intervals
Case 1: "Less Than"
\( |u| < c \implies -c < u < c \)
All solutions are between these values.
-c 0 +c
Case 2: "Greater Than"
\( |u| > c \implies u < -c \;\text{ OR }\; u > c \)
All solutions are outside of these values.
-c 0 +c
Example 1 \( 2|x + 1| + 5 \ge 9 \)
2|x + 1| + 5 ≥ 9 - 5 - 5
2|x + 1| 2
≥
4 2
|x + 1| ≥ 2
Positive Component
x + 1 ≥ 2 - 1 - 1
x ≥ 1
Negative Component
x + 1 ≤ -2 - 1 - 1
x ≤ -3
\( x \ge 1 \quad\text{OR}\quad x \le -3 \)
-3 0 1
Interval: \( (-\infty, -3] \cup [1, \infty) \)
Example 2 \( -3|x - 2| + 7 > -8 \)
-3|x - 2| + 7 > -8 - 7 - 7
-3|x - 2| -3
>
-15 -3
|x - 2| < 5
-5 < x - 2 < 5 + 2 + 2 + 2
-3 < x < 7
\( -3 < x < 7 \)
-3 2 7
Interval: \( (-3, 7) \)
Special Cases (When constant is negative after isolating)
\( |u| < \text{negative} \) \( \emptyset \) No Solution
\( |u| > \text{negative} \) \( (-\infty, \infty) \) All Real Numbers
Unit 2 - Absolute Value Function
Lesson 3 - Absolute Value Inequalities
Absolute Value Inequalities Practice Lesson Practice
Absolute Value Inequalities
Name:
For problems 1–3, circle either AND or OR for each inequality:
1. \( |x - 9| < 5 \)
AND
/
OR
2. \( |3x + 1| \ge 13 \)
AND
/
OR
3. \( -2|x + 5| \le -10 \)
AND
/
OR
Solve each inequality, graph the solution set on the number line, and express in interval notation:
4. \( |x + 6| \le 4 \)
Interval:
5. \( |2x - 5| > 9 \)
Interval:
6. \( 3|x - 2| + 4 < 19 \)
Interval:
7. \( -4|2x + 1| - 7 \le -23 \)
Interval:
Unit 2 - Absolute Value Function
Lesson 3 - Absolute Value Inequalities
Absolute Value Guided Notes Guided Notes
Absolute Value Inequalities
Name:
Date:
Determining Intervals: "AND" vs. "OR"
Fill in the missing rules and graphs below
Case 1: "Less Than" (< or ≤) Less thAND
\( |u| < c \implies \)
All solutions are between / inside the boundaries.
Graph the solution between \(-c\) and \(+c\):
-c 0 +c
Case 2: "Greater Than" (> or ≥) GreatOR
\( |u| > c \implies \) OR
All solutions are outside pointing in opposite directions.
Graph the rays extending outward:
-c 0 +c
Example 1 \( 2|x + 1| + 5 \ge 9 \) GreatOR (≥)
Step 1: Isolate the absolute value Subtract 5, divide by 2
2|x + 1| + 5 ≥ 9
|x + 1| ≥
Step 2: Split into two inequalities (OR)
Positive Case
x + 1 ≥ 2
x ≥
Negative Case (Flip!)
x + 1 ≤ -2
x ≤
Combined:
Step 3: Graph and express in interval notation
-3 0 1
Interval:
Example 2 \( -3|x - 2| + 7 > -8 \) Divide by negative!
Step 1: Isolate the absolute value Flip inequality sign!
-3|x - 2| + 7 > -8
|x - 2| sign
Step 2: Set up compound inequality (AND)
< x - 2 <
Add 2 to all three parts:
< x <
Solution:
Step 3: Graph and express in interval notation
-3 2 7
Interval:
Special Cases (Constant is negative after isolating)
Fill in: "No Solution" or "All Real Numbers"
\( |u| < \text{negative} \) Can distance be < negative?
Fill in solution
\( |u| > \text{negative} \) Can distance be > negative?
Fill in solution
Unit 2 - Absolute Value Function
Lesson 3 - Absolute Value Inequalities
Absolute Value Practice Key Teacher Answer Key Algebra 2 • Unit 3
Full Solutions & Graphs
Absolute Value Inequalities — Practice Key
Target Standard: HSA-CED.A.1, HSA-REI.B.3
Problems 1–3: Circle AND or OR for each inequality:
Correct answers highlighted
1. \( |x - 9| < 5 \)
AND
/
OR
"Less than" \(\rightarrow\) bounded (AND)
2. \( |3x + 1| \ge 13 \)
AND
/
OR
"Greater than" \(\rightarrow\) disjoint (OR)
3. \( -2|x + 5| \le -10 \)
AND
/
OR
Divide by \(-2\) flips to \(|x+5| \ge 5\) (OR)
Problems 4–7: Step-by-step solutions, graphs, and interval notations:
All intermediate steps & endpoints shown
4. \( |x + 6| \le 4 \) AND / Bounded
Setup: \(-4 \le x + 6 \le 4\)
Subtract 6 from all parts:
\(-4 - 6 \le x \le 4 - 6\)
\(-10 \le x \le -2\)
-10 -6 -2
Interval Notation: \( [-10, -2] \)
5. \( |2x - 5| > 9 \) OR / Disjoint
Left branch:
Right branch:
\(x < -2 \quad\text{OR}\quad x > 7\)
-2 2.5 7
Interval Notation: \( (-\infty, -2) \cup (7, \infty) \)
6. \( 3|x - 2| + 4 < 19 \) Isolate First • AND
Subtract 4: \(3|x - 2| < 15\) • Divide by 3: \(\mathbf{|x - 2| < 5}\)
Compound setup: \(-5 < x - 2 < 5\)
Add 2 to all parts: \(-5 + 2 < x < 5 + 2\)
\(-3 < x < 7\)
-3 2 7
Interval Notation: \( (-3, 7) \)
7. \( -4|2x + 1| - 7 \le -23 \) Flip Sign! • OR
Add 7: \(-4|2x+1| \le -16\) • Divide -4: \(\mathbf{|2x + 1| \ge 4}\)
\(2x + 1 \le -4 \implies \mathbf{x \le -\frac{5}{2}}\)
\(2x + 1 \ge 4 \implies \mathbf{x \ge \frac{3}{2}}\)
\(x \le -2.5 \quad\text{OR}\quad x \ge 1.5\)
-5/2 0 3/2
Interval Notation: \( (-\infty, -\frac{5}{2}] \cup [\frac{3}{2}, \infty) \)
Unit 2 - Absolute Value Function
Lesson 3 - Absolute Value Inequalities (Answer Key)
