# 6.8: Absolute Value Equations

**Basic**Created by: CK-12

**Practice**Absolute Value Equations

Suppose that a movie director has been told by a film studio that the length of the movie he's making must not differ from 120 minutes by more than 10 minutes. What would be the longest and shortest acceptable movies he could make? What absolute value equation could you set up to find out?

### Absolute Value Equations

Absolute value situations can also involve unknown variables. For example, suppose the distance from zero is 16. What two points can this represent?

Begin by writing an absolute value sentence to represent this situation.

\begin{align*}16=|n|, \ where \ n=the \ missing \ value\end{align*}

Which two numbers are 16 units from zero?

\begin{align*}n=16 \ or \ n=-16\end{align*}

Absolute value situations can also involve distances from points other than zero. We treat such cases as compound inequalities, separating the two independent equations and solving separately.

Let's solve the following absolute value problems:

- Solve for \begin{align*}x: |x-4|=5\end{align*}
x:|x−4|=5 .

This equation looks like the distance definition:

\begin{align*}distance=|x-y| \ or \ |y-x|\end{align*}

The distance is 5, and the value of \begin{align*}y\end{align*}

Visually, we can see the answers are –1 and 9.

Algebraically, we separate the two absolute value equations and solve.

\begin{align*}x-4=5 \ and \ x-4=-(5)\end{align*}

By solving each, the solutions become:

\begin{align*}x=9 \ and \ x=-1\end{align*}

- Solve \begin{align*}|2x-7|=6\end{align*}
|2x−7|=6 .

Begin by separating this into its separate equations.

\begin{align*}2x-7=6 \ and \ 2x-7=-6\end{align*}

Solve each equation independently.

\begin{align*}2x-7& =6 && \qquad 2x-7=-6\\
2x-7+7& =6+7 && \ 2x-7+7=-6+7\\
2x& =13 && \qquad \ \ \quad 2x=1\\
x &=\frac{13}{2} && \qquad \qquad x=\frac{1}{2}\end{align*}

#### Now, let's apply absolute value equations to a real-world situation and solve the following problem:

A company packs coffee beans in airtight bags. Each bag should weigh 16 ounces but it is hard to fill each bag to the exact weight. After being filled, each bag is weighed and if it is more than 0.25 ounces overweight or underweight, it is emptied and repacked. What are the lightest and heaviest acceptable bags?

The varying quantity is the weight of the bag of coffee beans. Choosing a letter to represent this quantity and writing an absolute value equation yields:

\begin{align*}|w-16|=0.25\end{align*}

This equation says that the distance away from 16 is equal to 0.25.

Separate and solve.

\begin{align*}w-16& =0.25 && w-16=-0.25\\
w& =16.25 && \qquad \ w=15.75\end{align*}

The lightest bag acceptable is 15.75 ounces and the heaviest bag accepted is 16.25 ounces.

### Examples

#### Example 1

Earlier, you were told that the length of a film by a certain movie director cannot differ from 120 minutes by more than 10 minutes. What would be the longest and shortest acceptable movies he could make? What absolute value equation could you set up to find out?

The quantity that varies is the length of the movie. Let's use \begin{align*}m\end{align*}

\begin{align*}\left|m-120\right|=10\end{align*}

This equation says that the distance away from 120 is 10.

Separate and solve the equation.

\begin{align*}m-120& =10 && m-120=-10\\
m& =130 && \qquad \ m=110\end{align*}

The shortest the film can be is 110 minutes and the longest the film can be is 130 minutes.

#### Example 2

Solve for \begin{align*}n\end{align*}

\begin{align*}|-3n+5|=17\end{align*}

Solving each one separately, we get:

\begin{align*}-3n+5& =17 & -3n+5 &=-17\\
-3n &= 12 & -3n &= -22\\
n &= -4 & n &= \frac{22}{3}
\end{align*}

### Review

In 1–12, solve the absolute value equations and interpret the results by graphing the solutions on a number line.

- \begin{align*}|7u|=77\end{align*}
|7u|=77 - \begin{align*}|x - 5| = 10\end{align*}
|x−5|=10 - \begin{align*}|5r-6|=9\end{align*}
|5r−6|=9 - \begin{align*}1=\frac{|6+5z|}{5}\end{align*}
1=|6+5z|5 - \begin{align*}|8x|=32\end{align*}
|8x|=32 - \begin{align*}|\frac{m}{8}|=1\end{align*}
|m8|=1 - \begin{align*}|x+2|=6\end{align*}
|x+2|=6 - \begin{align*}|5x-2|=3\end{align*}
|5x−2|=3 - \begin{align*}51=|1-5b|\end{align*}
51=|1−5b| - \begin{align*}8=3+|10y+5|\end{align*}
8=3+|10y+5| - \begin{align*}|4x-1|=19\end{align*}
|4x−1|=19 - \begin{align*}8|x+6|=-48\end{align*}
8|x+6|=−48

- A company manufactures rulers. Their 12-inch rulers pass quality control if they're within \begin{align*}\frac{1}{32}\end{align*}
132 inches of the ideal length. What is the longest and shortest ruler that can leave the factory?

**Mixed Review**

- A map has a scale of \begin{align*}2 \ inch=125 \ miles\end{align*}
2 inch=125 miles . How far apart would two cities be on the map if the actual distance is 945 miles? - Determine the domain and range: \begin{align*}\left \{(-9,0),(-6,0),(-4,0),(0,0),(3,0),(5,0)\right \}\end{align*}
{(−9,0),(−6,0),(−4,0),(0,0),(3,0),(5,0)} . - Is the relation in question #15 a function? Explain your reasoning.
- Consider the problem \begin{align*}3(2x-7)=100\end{align*}
3(2x−7)=100 . Lei says the first step to solving this equation is to use the Distributive Property to cancel the parentheses. Hough says the first step to solving this equation is to divide by 3. Who is right? Explain your answer. - Graph \begin{align*}4x+y=6\end{align*}
4x+y=6 using its intercepts. - Write \begin{align*}\frac{3}{30}\end{align*}
330 as a percent. Round to the nearest hundredth. - Simplify \begin{align*}-5\frac{2}{3} \div \frac{71}{8}\end{align*}
−523÷718 .

### Review (Answers)

To see the Review answers, open this PDF file and look for section 6.8.

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Here you'll learn how to find the solutions to absolute value equations.

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