# 6.7: Absolute Value

**Basic**Created by: CK-12

**Practice**Absolute Value

What if two people stood back-to-back and then walked in opposite directions? The first person walked 2 miles, while the second person walked 5 miles. How far apart would they be? Could you use a number line to represent this scenario? In this Concept, you'll review the meaning of absolute value and learn to find the distance between two numbers on a number line so that you can solve problems such as this one.

### Guidance

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 A

*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*}.

**Distance on the Number Line**

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 B

*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.

Check on the graph below that the length of the line between the points -5 and 8 is 13 units long:

#### Example C

*Find the distance between -12 and 3.*

**Solution:**

Use the definition of distance. Let \begin{align*}x=3\end{align*} and \begin{align*}y=-12\end{align*}.

\begin{align*}distance=|3-(-12)|=|3+12|=|15|=15\end{align*}

The absolute value of 15 is 15, so 3 and -12 are 15 units apart.

### Guided Practice

*Find the distance between 8 and -1.*

**Solution:**

We use the distance formula. Let \begin{align*}x=8\end{align*} and \begin{align*}y=-1\end{align*}.

\begin{align*}distance=|8-(-1)|=|8+1|=|9|=9\end{align*}

Notice that if we let \begin{align*}x=-1\end{align*} and \begin{align*}y=8\end{align*}, we get:

\begin{align*}distance=|-1-8|=|-9|=9\end{align*}.

So, it does not matter which number we pick for \begin{align*}x\end{align*} and \begin{align*}y\end{align*}. We will get the same answer.

### Practice

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

**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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Here you'll review the meaning of absolute value and learn how to find the distance between two numbers on a number line.