# 6.5: Absolute Value Equations

**At Grade**Created by: CK-12

## 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, \begin{align*}|-120| = 120\end{align*}
|−120|=120 because \begin{align*}-120\end{align*}−120 is \begin{align*}120\;\mathrm{units}\end{align*}120units 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) \begin{align*}a\end{align*}a and \begin{align*}b\end{align*}b , \begin{align*}| a - b | = | b - a |\end{align*}|a−b|=|b−a| . - Have students rethink the simple absolute value equations \begin{align*}|x| = 3\end{align*}
|x|=3 and \begin{align*}|x| = 10\end{align*}|x|=10 in Example 3 as \begin{align*}|?| = 3\end{align*}|?|=3 and \begin{align*}|?| = 10\end{align*}|?|=10 , respectively. (Which numbers are \begin{align*}3\;\mathrm{units}\end{align*}3units from the origin? \begin{align*}10\;\mathrm{units}\end{align*}10units from the origin?) - Have students interpret absolute value equations out loud. \begin{align*}|x - 2| = 7\end{align*}
|x−2|=7 means “those numbers on the number line \begin{align*}7\;\mathrm{units}\end{align*}7units away from \begin{align*}2\end{align*}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 \begin{align*}2\end{align*}
2 solutions. Have students consider absolute value equations with \begin{align*}1\end{align*}1 or solutions such as \begin{align*}|x - 2| = 0\end{align*}|x−2|=0 and \begin{align*}|x - 2| = 5\end{align*}|x−2|=5 , respectively.

Students have trouble reconciling the definition “\begin{align*}|x| = -x\end{align*}

- Let \begin{align*}x = -5\end{align*}
x=−5 . Then \begin{align*}|-5|=-(-5)\end{align*}|−5|=−(−5) since \begin{align*}-5\end{align*}−5 is negative (using the definition). This simplifies to \begin{align*}|-5| = 5\end{align*}|−5|=5 . -
\begin{align*}|-5|=5\end{align*}
|−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.

Additional Examples:

a. *Solve the equation and interpret the answer.*

\begin{align*} |x + 1| = 3\end{align*}

Solution: As it stands, the equation cannot be interpreted in terms of distance. Rewrite the equation with a minus sign: \begin{align*}|x - (-1)| = 3\end{align*}

b. *Solve the equation and interpret the answer.*

\begin{align*}|3x - 6| = 8\end{align*}

Hint: As it stands, the equation cannot be interpreted in terms of distance. Rewrite the equation by dividing both sides by \begin{align*}3\end{align*}

\begin{align*}\left |\frac{\cancel{3}x} {\cancel{3}} - \frac{6} {3}\right | & = \frac{8} {3}\\
|x - 2| & = \frac{8} {3}\end{align*}

This last equation can now be interpreted as those numbers \begin{align*}8/3\;\mathrm{units}\end{align*}

Treat the absolute value as a grouping symbol when appropriate.

- The distributive law holds in expressions such as \begin{align*}3|x - 4| = |3x - 12|\end{align*}
3|x−4|=|3x−12| and \begin{align*}\frac{1} {3} |3x - 6| = \left |\frac{\cancel{3}x} {\cancel{3}} - \frac{6} {3}\right | = |x - 2|\end{align*}13|3x−6|=∣∣∣3x3−63∣∣∣=|x−2| . - The distributive law does not hold in an expression such as \begin{align*}-2 |x - 3| \neq | -2x + 6 |\end{align*}
−2|x−3|≠|−2x+6| - In general, distribute into absolute value \begin{align*}a \cdot |b| = |a \cdot b|\end{align*}
a⋅|b|=|a⋅b| when \begin{align*}a = |a|\end{align*}a=|a| ; i.e., for positive numbers \begin{align*}a\end{align*}a . - These steps are based on the property \begin{align*}|a| \cdot |b| = |a \cdot b|\end{align*}
|a|⋅|b|=|a⋅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 \begin{align*}y = |x|\end{align*}

- The graph has a \begin{align*}``V''\end{align*} shape, consisting of two rays that meet at a sharp point, called the vertex or cusp.
- One side of the \begin{align*}``V''\end{align*} 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, \begin{align*}-2|x - 3| \neq | -2x + 6 |\end{align*}.

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

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