<meta http-equiv="refresh" content="1; url=/nojavascript/"> Absolute Value Equations | CK-12 Foundation
You are reading an older version of this FlexBook® textbook: CK-12 Algebra I Teacher's Edition Go to the latest version.

# 6.5: Absolute Value Equations

Created by: CK-12
0  0  0

## Learning Objectives

At the end of this lesson, students will be able to:

• Solve an absolute value equation.
• Analyze solutions to absolute value equations.
• Graph absolute value functions.
• Solve real-world problems using absolute value equations.

## Vocabulary

Terms introduced in this lesson:

absolute value
absolute value equations
vertex or cusp

## Teaching Strategies and Tips

Focus on the interpretation of absolute value as a distance.

• In Example 1b, $|-120| = 120$ because $â€“120$ is $120\;\mathrm{units}$ from the origin.
• In Example 2, have students explain, using a distance argument, why the order in which the two numbers are subtracted is not important. In general, for any two numbers (or points) $a$ and $b$, $| a - b | = | b - a |$.
• Have students rethink the simple absolute value equations $|x| = 3$ and $|x| = 10$ in Example 3 as $|?| = 3$ and $|?| = 10$, respectively. (Which numbers are $3\;\mathrm{units}$ from the origin? $10\;\mathrm{units}$ from the origin?)
• Have students interpret absolute value equations out loud. $|x - 2| = 7$ means “those numbers on the number line $7\;\mathrm{units}$ away from $2$.” See Examples 4-6. Encourage students to draw the number line and mark the possible solutions.
• Using the distance interpretation, point out that absolute value equations (involving only linear functions) can have no more than $2$ solutions. Have students consider absolute value equations with $1$ or $0$ solutions such as $|x - 2| = 0$ and $|x - 2| = 5$, respectively.

Students have trouble reconciling the definition “$|x| = -x$ if $x$ is negative” and the fact that “absolute value changes a negative number into its positive inverse.” Offer an example:

• Let $x = â€“5$. Then $|-5|=-(-5)$ since $â€“5$ is negative (using the definition). This simplifies to $|-5| = 5$.
• $|-5|=5$ since absolute value changes a negative number into its positive inverse.

Use Examples 5 and 6 to show students how to rewrite absolute value equations so that the distance interpretation is clearer.

a. Solve the equation and interpret the answer.

$|x + 1| = 3$

Solution: As it stands, the equation cannot be interpreted in terms of distance. Rewrite the equation with a minus sign: $|x - (-1)| = 3$ which can now be interpreted as those numbers $3\;\mathrm{units}$ away from $-1$. Therefore, the solution set is $\left \{2, -4\right \}$.

b. Solve the equation and interpret the answer.

$|3x - 6| = 8$

Hint: As it stands, the equation cannot be interpreted in terms of distance. Rewrite the equation by dividing both sides by $3$:

$\left |\frac{\cancel{3}x} {\cancel{3}} - \frac{6} {3}\right | & = \frac{8} {3}\\|x - 2| & = \frac{8} {3}$

This last equation can now be interpreted as those numbers $8/3\;\mathrm{units}$ away from $2$.

Treat the absolute value as a grouping symbol when appropriate.

• The distributive law holds in expressions such as $3|x - 4| = |3x - 12|$ and $\frac{1} {3} |3x - 6| = \left |\frac{\cancel{3}x} {\cancel{3}} - \frac{6} {3}\right | = |x - 2|$.
• The distributive law does not hold in an expression such as $-2 |x - 3| \neq | -2x + 6 |$
• In general, distribute into absolute value $a \cdot |b| = |a \cdot b|$ when $a = |a|$; i.e., for positive numbers $a$.
• These steps are based on the property $|a| \cdot |b| = |a \cdot b|$.

When beginning to graph absolute value functions, encourage students to make a table of values such as those in Examples 7 and 8.

Have students plot and describe in words the basic graph $y = |x|$. Allow them to observe the essential properties of the absolute value graph:

• The graph has a $â€œVâ€$ shape, consisting of two rays that meet at a sharp point, called the vertex or cusp.
• One side of the $â€œVâ€$ has positive slope and other side negative slope.
• The vertex is located at the point where the expression inside the absolute value is equal to zero.

## Error Troubleshooting

General Tip: Remind students not to distribute a negative into the absolute value expression. For example, $-2|x - 3| \neq | -2x + 6 |$.

General Tip: Students may misinterpret “absolute value is always positive” and commit the error $-|3-5|=2$. Suggest to students that in such situations $-|3 - 5| = (-1)|3 - 5|$. The multiplication by $-1$ happens after the absolute value has been performed, and so $-|3 - 5| = (-1)|3 - 5| = (-1) |-2| = (-1) 2 = -2$.

Feb 22, 2012

Aug 22, 2014