# 6.7: Absolute Value

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

**Practice**Absolute Value

What if you were given two points like -8 and 12? How could you find the distance between them on a number line? After completing this Concept, you'll be able to use absolute value properties to solve problems like this one.

### Watch This

CK-12 Foundation: 0607S Absolute Value Equations (H264)

### Guidance

Timmy is trying out his new roller skates. He’s not allowed to cross the street yet, so he skates back and forth in front of his house. If he skates 20 yards east and then 10 yards west, how far is he from where he started? What if he skates 20 yards west and then 10 yards east?

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 each a distance of 4 units away from zero.

\begin{align*}|4|\end{align*} represents the distance from 4 to zero, which equals 4.

\begin{align*}|-4|\end{align*} represents the distance from -4 to zero, which also equals 4.

In fact, for any real number \begin{align*}x\end{align*}:

\begin{align*}|x| = x\end{align*} if \begin{align*}x\end{align*} is not negative, and \begin{align*}|x| = -x\end{align*} if \begin{align*}x\end{align*} is negative.

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

#### Example A

*Evaluate the following absolute values.*

a) \begin{align*}|25|\end{align*}

b) \begin{align*}|-120|\end{align*}

c) \begin{align*}|-3|\end{align*}

d) \begin{align*}|55|\end{align*}

e) \begin{align*}\left | - \frac{5}{4} \right |\end{align*}

**Solution**

a) \begin{align*}|25|= 25\end{align*} Since 25 is a positive number, the absolute value does not change it.

b) \begin{align*}|-120| = 120\end{align*} Since -120 is a negative number, the absolute value makes it positive.

c) \begin{align*}|-3| = 3\end{align*} Since -3 is a negative number, the absolute value makes it positive.

d) \begin{align*}|55| = 55\end{align*} Since 55 is a positive number, the absolute value does not change it.

e) \begin{align*}\left | -\frac{5}{4} \right | =\frac{5}{4}\end{align*} Since \begin{align*}-\frac{5}{4}\end{align*} is a negative number, the absolute value makes it positive.

Absolute value is very useful in finding the distance between two points on the number line. The **distance** between any two points \begin{align*}a\end{align*} and \begin{align*}b\end{align*} on the number line is \begin{align*}|a - b|\end{align*} or \begin{align*}|b - a|\end{align*}.

For example, the distance from 3 to -1 on the number line is \begin{align*}|3 - (-1)| = |4| = 4\end{align*}.

We could have also found the distance by subtracting in the opposite order: \begin{align*}|-1 - 3| = |-4| = 4\end{align*}. This makes sense because the distance is the same whether you are going from 3 to -1 or from -1 to 3.

#### Example B

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

a) 6 and 15

b) -5 and 8

c) -3 and -12

**Solution**

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

a) \begin{align*}\text{distance} = |6 - 15| = |-9| = 9\end{align*}

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

c) \begin{align*}\text{distance} = |-3 - (-12)| = |9| = 9\end{align*}

**Remember:** When we computed the change in \begin{align*}x\end{align*} and the change in \begin{align*}y\end{align*} as part of the slope computation, these values were positive or negative, depending on the direction of movement. In this discussion, “distance” means a positive distance only.

**Solve an Absolute Value Equation**

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

\begin{align*}|x|=8\end{align*}

This means that the distance from the number \begin{align*}x\end{align*} to zero is 8. There are two numbers that satisfy this condition: 8 and -8.

When we solve absolute value equations we always consider 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 C

*Solve the following absolute value equations.*

a) \begin{align*}|x| = 3\end{align*}

b) \begin{align*}|x| = 10\end{align*}

**Solution**

a) There are two possibilities: **\begin{align*}x = 3\end{align*} and \begin{align*}x = -3.\end{align*}**

b) There are two possibilities: **\begin{align*}x = 10\end{align*} and \begin{align*}x = -10.\end{align*}**

Watch this video for help with the Examples above.

CK-12 Foundation: Absolute Value Equations

### Vocabulary

- The absolute value of a number is its distance from zero on a number line.
- \begin{align*}|x|=x\end{align*} if \begin{align*}x\end{align*} is not negative, and \begin{align*}|x|=-x\end{align*} if \begin{align*}x\end{align*} is negative.
- An equation or inequality with an absolute value in it
**splits into two equations,**one where the expression inside the absolute value sign is positive and one where it is negative. When the expression within the absolute value is**positive**, then the absolute value signs do nothing and can be omitted. When the expression within the absolute value is**negative,**then the expression within the absolute value signs must be negated before removing the signs. - Inequalities of the type \begin{align*}|x|<a\end{align*} can be rewritten as “\begin{align*}-a < x < a\end{align*}.”
- Inequalities of the type \begin{align*}|x|>b\end{align*} can be rewritten as “\begin{align*}x < -b\end{align*} or \begin{align*}x > b\end{align*}.”

### Guided Practice

*Find the distance between the values \begin{align*}-\frac{1}{3}\end{align*} and \begin{align*}\frac{1}{5}\end{align*} on the number line.*

**Solution:**

The distance is the absolute value of the difference:

\begin{align*}\left|-\frac{1}{3}-\frac{1}{5}\right|&= \quad \text{Set up the absolute value.} \\ \left|-\frac{5}{15}-\frac{3}{15}\right|&= \quad \text{Give the two terms common denominators.}\\ \left|\frac{-5-3}{15}\right|&= \quad \text{Combine the terms.}\\ \left|\frac{-8}{15}\right|&= \quad \text{Simplify.}\\ \frac{8}{15} &= \quad \text{Evaluate.} \end{align*}

The distance between the two points is \begin{align*}\frac{8}{15}\end{align*}.

### Practice

Evaluate the absolute values.

- \begin{align*}|250|\end{align*}
- \begin{align*}|-12|\end{align*}
- \begin{align*}|-0.003|\end{align*}
- \begin{align*}\left | -\frac{2}{5} \right |\end{align*}
- \begin{align*}\left | \frac{1}{10} \right |\end{align*}

Find the distance between the points.

- 12 and -11
- 5 and 22
- -9 and -18
- -2 and 3
- -0.012 and 1.067
- \begin{align*}-\frac{2}{3}\end{align*} and \begin{align*}\frac{7}{8}\end{align*}

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Here you'll learn how to find the distance between two values on a number line and solve equations involving absolute values.