6.7: 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.
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CK12 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*}
\begin{align*}4\end{align*}
In fact, for any real number \begin{align*}x\end{align*}
\begin{align*}x = x\end{align*}
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*}
b) \begin{align*}120 = 120\end{align*}
c) \begin{align*}3 = 3\end{align*}
d) \begin{align*}55 = 55\end{align*}
e) \begin{align*}\left  \frac{5}{4} \right  =\frac{5}{4}\end{align*}
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*}
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*}
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*}
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*}
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*}
b) There are two possibilities: \begin{align*}x = 10\end{align*}
Watch this video for help with the Examples above.
CK12 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*}
x=x if \begin{align*}x\end{align*}x is not negative, and \begin{align*}x=x\end{align*}x=−x if \begin{align*}x\end{align*}x 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*}
x<a can be rewritten as “\begin{align*}a < x < a\end{align*}−a<x<a .”  Inequalities of the type \begin{align*}x>b\end{align*}
x>b can be rewritten as “\begin{align*}x < b\end{align*}x<−b 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{53}{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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Description
Learning Objectives
Here you'll learn how to find the distance between two values on a number line and solve equations involving absolute values.