# 6.5: Absolute Value Equations

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

In Chapter 2, this textbook introduced the operation **absolute value**. The absolute value of a number is the distance from zero on a number line. The numbers 4 and –4 are each four units away from zero on a number line. So, \begin{align*}|4|=4\end{align*} *and* \begin{align*}|-4|=4\end{align*}.

Below is a more formal definition of absolute value.

For any real number \begin{align*}x\end{align*},

\begin{align*}|x|& =x \ for \ all \ x \ge 0\\ |x|& =-x (read \ \text{the opposite of} \ x) \ for \ all \ x<0\end{align*}

The second part of this definition states that the absolute value of a negative number is its opposite (a positive number).

**Example 1:** *Evaluate* \begin{align*}|-120|\end{align*}.

**Solution:** The absolute value of a negative number is its inverse, or opposite. Therefore, \begin{align*}|-120|=-(-120)=120\end{align*}.

Because the absolute value is always positive, it can be used to find the distance between two values on a number line.

The distance between two values \begin{align*}x\end{align*} and \begin{align*}y\end{align*} on a number line is found by:

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

**Example 2:** *Find the distance between –5 and 8.*

**Solution:** Use the definition of distance. Let \begin{align*}x=-5\end{align*} and \begin{align*}y=8\end{align*}.

\begin{align*}distance=|-5-8|=|-13|\end{align*}

The absolute value of –13 is 13, so –5 and 8 are 13 units apart.

## Solving an Absolute Value Equation

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 distance from points other than zero. We treat such cases as compound inequalities, separating the two independent equations and solving separately.

**Example 3:** *Solve for* \begin{align*}x: |x-4|=5\end{align*}.

**Solution:** 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*} is 4. We are looking for two values that are five units away from four on a number line.

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*}.

Begin by separating this into its separate equations.

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

Example: *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?*

Solution: 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*}

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.

## Graphing an Absolute Value Equation

Absolute value equations can be graphed in a way that is similar to graphing linear equations. By making a table of values, you can get a clear picture of what an absolute value equation will look like.

Example: *Graph the solutions to* \begin{align*}y=|x-1|\end{align*}.

Solution: Make a table of values and plot the ordered pairs.

\begin{align*}x\end{align*} | \begin{align*}y=|x-1|\end{align*} |
---|---|

–2 | \begin{align*}|-2-1|=3\end{align*} |

–1 | \begin{align*} |-1-1|=2\end{align*} |

0 | \begin{align*} |0-1|=1\end{align*} |

1 | \begin{align*} |1-1|=0\end{align*} |

2 | \begin{align*} |2-1|=1\end{align*} |

3 | \begin{align*} |3-1|=2\end{align*} |

Every absolute value graph will make a “V”-shaped figure. It consists of two pieces: one with a negative slope and one with a positive slope. The point of their intersection is called the **vertex**. An absolute value graph is **symmetrical**, meaning it can be folded in half on its **line of symmetry**.

Absolute value equations can always be graphed by making a table of values. However, you can use the vertex and symmetry to help shorten the graphing process.

Step 1: Find the vertex by determining which value of \begin{align*}x\end{align*} makes the distance zero.

Step 2: Using this value as the center of the \begin{align*}x-\end{align*}values, choose several values greater than this value and several values less than this value.

Example: Graph \begin{align*}y=|x+5|\end{align*}.

Solution: Determine which \begin{align*}x-\end{align*}value equals a distance of zero.

\begin{align*}0& =|x+5|\\ x& =-5\end{align*}

Therefore, (–5, 0) is the vertex of the graph and represents the center of the table of values.

Create the table and plot the ordered pairs.

\begin{align*}x\end{align*} | \begin{align*}y=|x+5|\end{align*} |
---|---|

–7 | \begin{align*} |-7+5|=2\end{align*} |

–6 | \begin{align*} |-6+5|=1\end{align*} |

–5 | \begin{align*} |-5+5|=0\end{align*} |

–4 | \begin{align*} |-4+5|=1\end{align*} |

–3 | \begin{align*} |-3+5|=2\end{align*} |

## Practice Set

Sample explanations for some of the practice exercises below are available by viewing the following video. Note that there is not always a match between the number of the practice exercise in the video and the number of the practice exercise listed in the following exercise set. However, the practice exercise is the same in both.

CK-12 Basic Algebra: Absolute Value Equations (10:41)

Evaluate the absolute value.

- \begin{align*}|250|\end{align*}
- \begin{align*}|-12|\end{align*}
- \begin{align*}|-\frac{2}{5}|\end{align*}
- \begin{align*}|\frac{1}{10}|\end{align*}

Find the distance between the points.

- 12 and –11
- 5 and 22
- –9 and –18
- –2 and 3
- \begin{align*}\frac{2}{3}\end{align*} and –11
- –10.5 and –9.75
- 36 and 14

In 12 – 22, solve the absolute value equations and interpret the results by graphing the solutions on the number line.

- \begin{align*}|x - 5| = 10\end{align*}
- \begin{align*}|5r-6|=9\end{align*}
- \begin{align*}1=\frac{|6+5z|}{5}\end{align*}
- \begin{align*}|8x|=32\end{align*}
- \begin{align*}|\frac{m}{8}|=1\end{align*}
- \begin{align*}|x+2|=6\end{align*}
- \begin{align*}|5x-2|=3\end{align*}
- \begin{align*}51=|1-5b|\end{align*}
- \begin{align*}8=3+|10y+5|\end{align*}
- \begin{align*}|4x-1|=19\end{align*}
- \begin{align*}8|x+6|=-48\end{align*}
- What two features of an absolute value graph help determine the appropriate \begin{align*}x-\end{align*}values to use for a table?
- The vertex of an absolute value equation is (0.5, 0). Give several \begin{align*}x-\end{align*}values that would be appropriate for a table.

In 25 – 35, graph the function.

- \begin{align*}y=|x+3|\end{align*}
- \begin{align*}y=|x-6|\end{align*}
- \begin{align*}y=|4x+2|\end{align*}
- \begin{align*}y=\left |\frac{x}{3}-4\right |\end{align*}
- \begin{align*}|x-4|=y\end{align*}
- \begin{align*}-|x-2|=y\end{align*}
- \begin{align*}y=|x|-2\end{align*}
- \begin{align*}y=|x|+3\end{align*}
- \begin{align*}y=\frac{1}{2} |x|\end{align*}
- \begin{align*}y=4|x|-2\end{align*}
- \begin{align*}y=\left |\frac{1}{2} x\right |+6\end{align*}
- A company manufactures rulers. Their 12-inch rulers pass quality control if they within \begin{align*}\frac{1}{32}\end{align*} inches of the ideal length. What is the longest and shortest ruler that can leave the factory?

**Mixed Review**

- Solve: \begin{align*}6t-14<2t+7\end{align*}.
- The speed limit of a semi-truck on the highway is between 45 mph and 65 mph.
- Write this situation as a compound inequality
- Graph the solutions on a number line.

- Lloyd can only afford transportation costs of less than $276 per month. His monthly car payment is $181 and he sets aside $25 per month for oil changes and other maintenance costs. How much can he afford for gas?
- Simplify \begin{align*}\sqrt{12} \times \sqrt{3}\end{align*}.
- A hush puppy recipe calls for 3.4 ounces of flour for one batch of 8 hush puppies. You need to make 56 hush puppies. How much flour do you need?
- What is the additive inverse of 124?
- What is the multiplicative inverse of 14?
- Define the
*Addition Property of Equality*.

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