6.8: Absolute Value Equations
What if you were asked to solve an absolute value equation like \begin{align*}3x4=5\end{align*}
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CK12 Foundation: 0608S Analyze Solutions to Absolute Value Equations (H264)
Guidance
In the previous concept, we saw how to solve simple absolute value equations. In this concept, you will see how to solve more complicated absolute value equations.
Example A
Solve the equation \begin{align*}x4=5\end{align*}
Solution
We consider two possibilities: the expression inside the absolute value sign is nonnegative or is negative. Then we solve each equation separately.
\begin{align*}& x4 = 5 \quad \text{and} \quad x4=5\\
& \quad \ \ x=9 \qquad \qquad \quad \ x=1\end{align*}
\begin{align*}x = 9\end{align*}
The equation \begin{align*}x4=5\end{align*}
Example B
Solve the equation \begin{align*}x+3=2\end{align*}
Solution
Solve the two equations:
\begin{align*}& x+3 = 2 \quad \text{and} \quad \ \ x+3=2\\
& \quad \ \ x=1 \qquad \qquad \quad \ x=5\end{align*}
\begin{align*}x = 5\end{align*}
The equation \begin{align*}x+3=2\end{align*}
Solve RealWorld Problems Using Absolute Value Equations
Example C
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; 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 weight of each bag is allowed to be 0.25 ounces away from 16 ounces; in other words, the difference between the bag’s weight and 16 ounces is allowed to be 0.25 ounces. So if \begin{align*}x\end{align*}
Now we must consider the positive and negative options and solve each equation separately:
\begin{align*}& x16 = 0.25 \qquad \text{and} \quad x16 =0.25\\
& \qquad x=16.25 \qquad \qquad \qquad \ \ x=15.75\end{align*}
The lightest acceptable bag weighs 15.75 ounces and the heaviest weighs 16.25 ounces.
We see that \begin{align*}16.25  16 = 0.25 \ ounces\end{align*}
The answer checks out.
The answer you just found describes the lightest and heaviest acceptable bags of coffee beans. But how do we describe the total possible range of acceptable weights? That’s where inequalities become useful once again.
Watch this video for help with the Examples above.
CK12 Foundation: Analyze Solutions to 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*}x>b .”
Guided Practice
Solve the equation \begin{align*}2x7=6\end{align*}
Solution
Solve the two equations:
\begin{align*}& 2x7 = 6 \qquad \qquad \quad 2x7=6\\
& \quad \ \ 2x=13 \qquad \text{and} \qquad \ \ 2x=1\\
& \quad \ \ \ x=\frac{13}{2} \qquad \qquad \qquad \ \ x=\frac{1}{2}\end{align*}
Answer: \begin{align*}x=\frac{13}{2}\end{align*}
The interpretation of this problem is clearer if the equation \begin{align*}2x7=6\end{align*}
Practice
Solve the absolute value equations and interpret the results by graphing the solutions on the number line.

\begin{align*}x5=10\end{align*}
x−5=10 
\begin{align*}x+2=6\end{align*}
x+2=6 
\begin{align*}5x2=3\end{align*}
5x−2=3 
\begin{align*}x4=3\end{align*}
x−4=−3 
\begin{align*}\left2x\frac{1}{2}\right=10\end{align*}
∣∣∣2x−12∣∣∣=10 
\begin{align*}x+5=\frac{1}{5}\end{align*}
−x+5=15 
\begin{align*}\left\frac{1}{2}x5\right=100\end{align*}
∣∣∣12x−5∣∣∣=100 
\begin{align*}10x5=15\end{align*}
10x−5=15 
\begin{align*}0.1x+3=0.015\end{align*}
0.1x+3=0.015 
\begin{align*}272x=3x+2\end{align*}
27−2x=3x+2
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Here you'll learn how to how to solve more complicated absolute value equations and interpret your answers.