Learn on PengienVision, Algebra 1Chapter 1: Solving Equations and Inequalities

Lesson 7: Absolute Value Equations and Inequalities

In this Grade 11 enVision Algebra 1 lesson, students learn to write and solve absolute value equations and inequalities, including recognizing when an equation has no solution and rewriting absolute value inequalities as equivalent compound inequalities using "and" or "or" conditions. The lesson covers isolating absolute value expressions, splitting equations into two cases, and applying these skills to real-world problems involving distance and acceptable ranges.

Section 1

Absolute Value Equations

Property

For any algebraic expression, $u$ , and any positive real number, $a$ , if $|u| = a$ , then $u = -a$ or $u = a$ .
To solve an absolute value equation, first isolate the absolute value expression.
Then, write two equivalent equations and solve each one separately.

Examples

To solve $|3x - 1| - 5 = 10$ , first isolate the absolute value: $|3x - 1| = 15$ . Then set up two equations: $3x - 1 = 15$ or $3x - 1 = -15$ . Solving gives $x = \frac{16}{3}$ or $x = -\frac{14}{3}$ .
The equation $3|x - 8| + 11 = 5$ simplifies to $3|x - 8| = -6$ , or $|x - 8| = -2$ . Since an absolute value cannot be negative, there is no solution.
To solve $|x + 4| = |2x - 2|$ , set up two cases: $x + 4 = 2x - 2$ or $x + 4 = -(2x - 2)$ . Solving these yields $x = 6$ or $x = -\frac{2}{3}$ .

Explanation

If the absolute value of something is a positive number $a$ , it means the 'something' inside is either $a$ units to the right of zero or $a$ units to the left. This is why you must solve two separate cases.

Section 2

No Solution Case for Absolute Value Equations

Property

If $|ax + b| = c$ where $c < 0$ , then the equation has no solution because absolute value expressions are always non-negative: $|expression| \geq 0$ .

Examples

Section 3

Solving Absolute Value Less-Than Inequalities

Property

When an isolated absolute value inequality uses a "less than" ( $<$ ) or "less than or equal to" ( $\leq$ ) symbol with a positive number, it translates into a compound "AND" inequality.
If $|u| < a$ , the equivalent compound inequality is $-a < u < a$ .
This means the expression inside must be trapped between the negative and positive boundaries of that distance.

Examples

Simple Less-Than: Solve $|x| < 9$ .

You are looking for all numbers whose distance from zero is less than 9. Rewrite as a compound inequality: $-9 < x < 9$ . The solution is $(-9, 9)$ .

Multi-Step Less-Than: Solve $|2x - 5| \leq 3$ .

Rewrite as a compound "and" inequality: $-3 \leq 2x - 5 \leq 3$ .
Add 5 to all three parts: $2 \leq 2x \leq 8$ .
Divide all three parts by 2: $1 \leq x \leq 4$ . The solution is $[1, 4]$ .

Explanation

When an absolute value is "less than" a number, it means the expression is constrained. It is kept close to the center point. To solve it algebraically, you drop the absolute value bars and sandwich the inner expression between the negative and positive limits. You then solve all three parts of the chain at the same time to find the exact overlapping region.

Lesson overview

Expand to review the lesson summary and core properties.

Expand

Section 1

Absolute Value Equations

Property

Examples

To solve $|3x - 1| - 5 = 10$ , first isolate the absolute value: $|3x - 1| = 15$ . Then set up two equations: $3x - 1 = 15$ or $3x - 1 = -15$ . Solving gives $x = \frac{16}{3}$ or $x = -\frac{14}{3}$ .
The equation $3|x - 8| + 11 = 5$ simplifies to $3|x - 8| = -6$ , or $|x - 8| = -2$ . Since an absolute value cannot be negative, there is no solution.
To solve $|x + 4| = |2x - 2|$ , set up two cases: $x + 4 = 2x - 2$ or $x + 4 = -(2x - 2)$ . Solving these yields $x = 6$ or $x = -\frac{2}{3}$ .

Explanation

Section 2

No Solution Case for Absolute Value Equations

Property

If $|ax + b| = c$ where $c < 0$ , then the equation has no solution because absolute value expressions are always non-negative: $|expression| \geq 0$ .

Examples

Section 3

Solving Absolute Value Less-Than Inequalities

Property

Examples

Simple Less-Than: Solve $|x| < 9$ .

You are looking for all numbers whose distance from zero is less than 9. Rewrite as a compound inequality: $-9 < x < 9$ . The solution is $(-9, 9)$ .

Multi-Step Less-Than: Solve $|2x - 5| \leq 3$ .

Rewrite as a compound "and" inequality: $-3 \leq 2x - 5 \leq 3$ .
Add 5 to all three parts: $2 \leq 2x \leq 8$ .
Divide all three parts by 2: $1 \leq x \leq 4$ . The solution is $[1, 4]$ .

Explanation

Overview

Lesson overview

Review the lesson summary and core properties without leaving the current page.

Overview

Section 1

Absolute Value Equations

Property

Examples

To solve $|3x - 1| - 5 = 10$ , first isolate the absolute value: $|3x - 1| = 15$ . Then set up two equations: $3x - 1 = 15$ or $3x - 1 = -15$ . Solving gives $x = \frac{16}{3}$ or $x = -\frac{14}{3}$ .
The equation $3|x - 8| + 11 = 5$ simplifies to $3|x - 8| = -6$ , or $|x - 8| = -2$ . Since an absolute value cannot be negative, there is no solution.
To solve $|x + 4| = |2x - 2|$ , set up two cases: $x + 4 = 2x - 2$ or $x + 4 = -(2x - 2)$ . Solving these yields $x = 6$ or $x = -\frac{2}{3}$ .

Explanation

Section 2

No Solution Case for Absolute Value Equations

Property

If $|ax + b| = c$ where $c < 0$ , then the equation has no solution because absolute value expressions are always non-negative: $|expression| \geq 0$ .

Examples

Section 3

Solving Absolute Value Less-Than Inequalities

Property

Examples

Simple Less-Than: Solve $|x| < 9$ .

You are looking for all numbers whose distance from zero is less than 9. Rewrite as a compound inequality: $-9 < x < 9$ . The solution is $(-9, 9)$ .

Multi-Step Less-Than: Solve $|2x - 5| \leq 3$ .

Rewrite as a compound "and" inequality: $-3 \leq 2x - 5 \leq 3$ .
Add 5 to all three parts: $2 \leq 2x \leq 8$ .
Divide all three parts by 2: $1 \leq x \leq 4$ . The solution is $[1, 4]$ .