# 6.5: Absolute Value Equations

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

## Learning Objectives

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

## Introduction

The **absolute value** of a number is its distance from zero on a number line. There are always two numbers on the number line that are the same distance from zero. For instance, the numbers 4 and -4 are both a distance of 4 units away from zero.

In fact, for any real number

Absolute value has no effect on a positive number but changes a negative number into its positive inverse.

**Example 1**

*Evaluate the following absolute values.*

a)

b)

c)

d)

e)

**Solution:**

a)

b)

c)

d)

e)

Absolute value is very useful in finding the distance between two points on the number line. The distance between any two points

For example, the distance from 3 to -1 on the number line is

We could have also found the distance by subtracting in the reverse order,

This makes sense because the distance is the same whether you are going from 3 to -1 or from -1 to 3.

**Example 2**

*Find the distance between the following points on the number line.*

a) 6 and 15

b) -5 and 8

c) -3 and -12

**Solutions**

Distance is the absolute value of the difference between the two points.

a) Distance

b) Distance

c) Distance

**Remember:** When we computed the change in

## Solve an Absolute Value Equation

We now want to solve equations involving absolute values. Consider the following equation.

This means that the distance from the number

When we solve absolute value equations we always consider 4 two possibilities.

- The expression inside the absolute value sign is not negative.
- The expression inside the absolute value sign is negative.

Then we solve each equation separately.

**Example 3**

*Solve the following absolute value equations.*

a)

b)

**Solution**

a) There are two possibilities

b) There are two possibilities

## Analyze Solutions to Absolute Value Equations

**Example 4**

*Solve the equation and interpret the answers.*

**Solution**

We consider two possibilities. The expression inside the absolute value sign is not negative or is negative. Then we solve each equation separately.

**Answer**

Equation

**Example 5**

*Solve the equation ∣x+3∣=2 and interpret the answers.*

**Solution**

Solve the two equations.

**Answer**

Equation

**Example 6**

*Solve the equation ∣2x−7∣=6 and interpret the answers.*

**Solution**

Solve the two equations.

**Answer**

The interpretation of this problem is clearer if the equation \begin{align*}\mid 2x-7 \mid =6 \end{align*} was divided by 2 on both sides. We obtain \begin{align*} \big | x -\frac{7}{2} \big | =3 \end{align*}. The question is “What numbers on the number line are 3 units away from \begin{align*} \frac{7}{2}\end{align*}?” There are two possibilities \begin{align*} \frac{13}{2}\end{align*} and \begin{align*} \frac{1}{2}\end{align*}.

## Graph Absolute Value Functions

You will now learn how to graph absolute value functions. Consider the function:

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

Let’s graph this function by making a table of values.

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

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

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

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

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

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

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

4 | \begin{align*}y=\mid 4-1\mid = \mid 3 \mid =3\end{align*} |

You can see that the graph of an absolute value function makes a big “V”. It consists of two line rays (or line segments), one with positive slope and one with negative slope joined at the **vertex** or **cusp**.

We saw in previous sections that to solve an absolute value equation we need to consider two options.

- The expression inside the absolute value is not negative.
- The expression inside the absolute value is negative.

The graph of \begin{align*}y= \mid x -1\mid\end{align*} is a combination of two graphs.

*Option 1*

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

when \begin{align*}x-1 \geq 0\end{align*}

*Option 2*

\begin{align*}y=-(x-1)\end{align*} or \begin{align*}y=-x+1\end{align*}

when \begin{align*} x-1 <0\end{align*}

These are both graphs of straight lines.

The two straight lines meet at the vertex. We find the vertex by setting the expression inside the absolute value equal to zero.

\begin{align*}x-1=0\end{align*} or \begin{align*}x=1\end{align*}

We can always graph an absolute value function using a table of values. However, we usually use a simpler procedure.

*Step 1* Find the vertex of the graph by setting the expression inside the absolute value equal to zero and solve for \begin{align*}x\end{align*}.

*Step 2* Make a table of values that includes the vertex, a value smaller than the vertex and a value larger than the vertex. Calculate the values of \begin{align*}y\end{align*} using the equation of the function.

*Step 3* Plot the points and connect with two straight lines that meet at the vertex.

**Example 7**

*Graph the absolute value function:* \begin{align*}y=\mid x+5\mid \end{align*}.

**Solution**

*Step 1* Find the vertex \begin{align*}x+5=0\end{align*} or \begin{align*}x=-5\end{align*} vertex.

*Step 2* Make a table of values.

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

\begin{align*}-8\end{align*} | \begin{align*}y=\mid -8+5 \mid = \mid -3 \mid =3\end{align*} |

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

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

*Step 3* Plot the points and draw two straight lines that meet at the vertex.

**Example 8**

*Graph the absolute value function* \begin{align*}y=\mid 3x -12 \mid\end{align*}.

**Solution**

*Step 1* Find the vertex \begin{align*}3x- 12 = 0\end{align*} so \begin{align*}x=4\end{align*} is the vertex.

*Step 2* Make a table of values:

\begin{align*}x\end{align*} | \begin{align*}y=\mid 3x-12\mid\end{align*} |
---|---|

0 | \begin{align*}y=\mid 3(0)-12 \mid = \mid -12\mid =12\end{align*} |

4 | \begin{align*}y=\mid 3(4)-12\mid = \mid 0 \mid =0\end{align*} |

8 | \begin{align*}y=\mid 3(8)-12 \mid = \mid 12 \mid =12\end{align*} |

*Step 3* Plot the points and draw two straight lines that meet at the vertex.

## Solve Real-World Problems Using Absolute Value Equations

**Example 9**

*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**

*Step 1*

We know that each bag should weigh 16 ounces.

A bag can weigh 0.25 ounces more or less than 16 ounces.

We need to find the lightest and heaviest bags that are acceptable.

Let \begin{align*}x =\end{align*} weight of the coffee bag in ounces.

*Step 2*

The equation that describes this problem is written as \begin{align*} \mid x -16 \mid 0.25 \end{align*}.

*Step 3*

Consider the positive and negative options and solve each equation separately.

\begin{align*}x-16&=0.25 & & & & x-16 =-0.25\\ & & \text{and} & & \\ x&=16.25 & & & & \qquad \ x=15.75\end{align*}

**Answer** The lightest acceptable bag weighs 15.75 ounces and the heaviest weighs 16.25 ounces.

*Step 4*

We see that \begin{align*}16.25 - 16 = 0.25\end{align*} ounces and \begin{align*}16 - 15.75 = 0.25\end{align*} ounces. The answers are 0.25 ounces bigger and smaller than 16 ounces respectively.

**The answer checks out.**

## Lesson Summary

- The absolute value of a number is its distance from zero on a number line.

\begin{align*}\mid x \mid =x \end{align*} if \begin{align*}x\end{align*} is not negative.

\begin{align*}\mid x \mid =-x \end{align*} if \begin{align*}x\end{align*} is negative.

- An equation with an absolute value in it
**splits into two equations.**

- The expression within the absolute value is
**positive,**then the absolute value signs do nothing and can be omitted. - The expression within the absolute value is
**negative,**then the expression within the absolute value signs must be negated before removing the signs.

## Review Questions

Evaluate the absolute values.

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

Find the distance between the points.

- 12 and -11
- 5 and 22
- -9 and -18
- -2 and 3

Solve the absolute value equations and interpret the results by graphing the solutions on the number line.

- \begin{align*} \mid x -5 \mid =10 \end{align*}
- \begin{align*} \mid x +2 \mid =6 \end{align*}
- \begin{align*} \mid 5x-2 \mid =3\end{align*}
- \begin{align*} \mid 4x-1 \mid =19\end{align*}

Graph the absolute value functions.

- \begin{align*} y =\mid x +3 \mid \end{align*}
- \begin{align*} y = \mid x -6 \mid \end{align*}
- \begin{align*} y = \mid 4x+2 \mid \end{align*}
- \begin{align*} y = \mid \frac{x}{3}-4\mid \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?

## Review Answers

- 250
- 12
- \begin{align*} \frac{2}{5}\end{align*}
- \begin{align*}\frac{1}{10}\end{align*}
- 23
- 17
- 9
- 5
- 15 and -5
- 4 and -8
- 1 and \begin{align*}-\frac{1}{5}\end{align*}
- 5 and \begin{align*}-\frac{9}{2}\end{align*}
- \begin{align*} 11 \frac{31} {32}\end{align*} and \begin{align*}12\frac{1} {32}\end{align*}

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