Absolute Value Equations and Inequalities (Chapter 2.5) – College Algebra Study Notes

Absolute Value Equations and Inequalities

Introduction

This section explores the properties, graphs, and solution methods for absolute value equations and inequalities. Mastery of these concepts is essential for understanding a wide range of algebraic and real-world problems.

Absolute Value Functions

Definition and Properties

• Absolute Value: The absolute value of a real number x, denoted $|x|$, is its distance from zero on the number line. It is always non-negative.

• Mathematical Definition:

  • $|x| = x$ if $x \geq 0$

  • $|x| = -x$ if $x < 0$

• Graph of $y = |x|$: The graph is a "V" shape with its vertex at the origin (0,0).

Graphing Absolute Value Functions

• To graph $y = |f(x)|$, reflect the portion of $y = f(x)$ that is below the x-axis across the x-axis.

• For $y = |x-2|$, the graph is the standard absolute value graph shifted right by 2 units.

• For $y = |1-2x|$, the graph is reflected and stretched compared to $y = |x|$.

Example: Graph $y = x-2$ and $y = |x-2|$. The graph of $y = |x-2|$ is identical to $y = x-2$ for $x \geq 2$, but for $x < 2$, it is reflected above the x-axis.

Example: Graph $y = 1-2x$ and $y = |1-2x|$. The graph of $y = |1-2x|$ is a reflection of $y = 1-2x$ above the x-axis wherever $1-2x$ is negative.

Solving Absolute Value Equations

General Approach

• An equation of the form $|A| = B$ (where $B \geq 0$) has two solutions: $A = B$ or $A = -B$.

• If $B < 0$, there is no solution because absolute value cannot be negative.

Example 1

Solve $\left| \frac{1}{2}x + 1 \right| = 5$

• Set up two equations:

  • $\frac{1}{2}x + 1 = 5$

  • $\frac{1}{2}x + 1 = -5$

• Solving:

  • $\frac{1}{2}x = 4 \implies x = 8$

  • $\frac{1}{2}x = -6 \implies x = -12$

• Solution set: $\{-12, 8\}$

Example 2

Solve $|5 + 3x| = -4$

• Since the right side is negative, there is no solution.

Example 3

Solve $|1-2x|-3 = -1$

• Add 3 to both sides: $|1-2x| = 2$

• Set up two equations:

  • $1-2x = 2 \implies x = -\frac{1}{2}$

  • $1-2x = -2 \implies x = \frac{3}{2}$

• Solution set: $\left\{-\frac{1}{2}, \frac{3}{2}\right\}$

Example 4

Solve $|4-2x| = 8$ graphically, numerically, and symbolically.

• Graphical Solution: Plot $y_1 = |4-2x|$ and $y_2 = 8$; the intersection points give the solutions.

• Numerical Solution: Use a table of values to find $x$ where $|4-2x| = 8$.

• Symbolic Solution:

  • $4-2x = 8 \implies x = -2$

  • $4-2x = -8 \implies x = 6$

• Solution set: $\{-2, 6\}$

Solving Absolute Value Inequalities

General Approach

• For $|A| < B$ (where $B > 0$): $-B < A < B$

• For $|A| > B$ (where $B > 0$): $A < -B$ or $A > B$

• If $B < 0$, $|A| < B$ has no solution; $|A| > B$ is always true for all real numbers except $A = 0$ if $B = 0$.

Example 5

The diameter $D$ in inches of a cylindrical can must satisfy $|D-3| \leq 0.02$. Find the maximum and minimum acceptable values for $D$.

• Set up: $-0.02 \leq D-3 \leq 0.02$

• Add 3 to all parts: $2.98 \leq D \leq 3.02$

• Maximum value: 3.02, Minimum value: 2.98

Example 6

Solve $|4-x| < 2$

• Set up: $-2 < 4-x < 2$

• Subtract 4: $-6 < -x < -2$

• Multiply by -1 (reverse inequalities): $6 > x > 2$ or $2 < x < 6$

• Solution set: $(2, 6)$

Example 7

Solve $|3x+2| > 1$

• Set up: $3x+2 > 1$ or $3x+2 < -1$

• $3x > -1 \implies x > -\frac{1}{3}$

• $3x < -3 \implies x < -1$

• Solution set: $(-\infty, -1) \cup (\frac{-1}{3}, \infty)$

Example 8

The thickness $A$ of a phone is about 9.33 mm, but actual phones may vary by not more than 0.02 mm. Write and solve an absolute value statement for this situation.

• Statement: $|A - 9.33| < 0.02$

• Set up: $-0.02 < A - 9.33 < 0.02$

• Add 9.33: $9.31 < A < 9.35$

• Interpretation: The actual thickness must be between 9.31 mm and 9.35 mm.

Example 9

Solve $|3-2x| < 5$

• Set up: $-5 < 3-2x < 5$

• Subtract 3: $-8 < -2x < 2$

• Divide by -2 (reverse inequalities): $4 > x > -1$ or $-1 < x < 4$

• Solution set: $(-1, 4)$

Example 10

Solve $|-3x+1| > 8$

• Set up: $-3x+1 > 8$ or $-3x+1 < -8$

• $-3x > 7 \implies x < -\frac{7}{3}$

• $-3x < -9 \implies x > 3$

• Solution set: $(-\infty, -\frac{7}{3}) \cup (3, \infty)$

Summary Table: Absolute Value Equations and Inequalities

Form	Solution Method	Number of Solutions
$|A| = B$ ($B > 0$)	$A = B$ or $A = -B$	2 (unless $B = 0$, then 1)
$|A| = B$ ($B < 0$)	No solution	0
$|A| < B$ ($B > 0$)	$-B < A < B$	Infinite (interval)
$|A| > B$ ($B > 0$)	$A < -B$ or $A > B$	Infinite (union of intervals)

Additional info: The above notes expand on the examples and methods shown in the slides, providing full algebraic steps, definitions, and interval notation for solution sets.