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# 6.5: Absolute Value Equations

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

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

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

\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*} which can now be interpreted as those numbers \begin{align*}3\;\mathrm{units}\end{align*} away from \begin{align*}-1\end{align*}. Therefore, the solution set is \begin{align*}\left \{2, -4\right \}\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*} away from \begin{align*}2\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*} and \begin{align*}\frac{1} {3} |3x - 6| = \left |\frac{\cancel{3}x} {\cancel{3}} - \frac{6} {3}\right | = |x - 2|\end{align*}.
• The distributive law does not hold in an expression such as \begin{align*}-2 |x - 3| \neq | -2x + 6 |\end{align*}
• In general, distribute into absolute value \begin{align*}a \cdot |b| = |a \cdot b|\end{align*} when \begin{align*}a = |a|\end{align*}; i.e., for positive numbers \begin{align*}a\end{align*}.
• These steps are based on the property \begin{align*}|a| \cdot |b| = |a \cdot b|\end{align*}.

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*}. Allow them to observe the essential properties of the absolute value graph:

• 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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Date Created:
Feb 22, 2012