<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
Skip Navigation
Our Terms of Use (click here to view) have changed. By continuing to use this site, you are agreeing to our new Terms of Use.

6.8: Absolute Value Equations

Difficulty Level: At Grade Created by: CK-12
Atoms Practice
Estimated11 minsto complete
Practice Absolute Value Equations
Estimated11 minsto complete
Practice Now
Turn In

Absolute Value Equations 

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.

Solving Absolute Value Equations 

1. Solve the equation \begin{align*}|x-4|=5\end{align*} and interpret the answers.

We consider two possibilities: the expression inside the absolute value sign is non-negative or is negative. Then we solve each equation separately.

\begin{align*}& x-4 = 5 \quad \text{and} \quad x-4=-5\\ & \quad \ \ x=9 \qquad \qquad \quad \ x=-1\end{align*}

\begin{align*}x = 9\end{align*} and \begin{align*}x = -1\end{align*} are the solutions.

The equation \begin{align*}|x-4|=5\end{align*} can be interpreted as “what numbers on the number line are 5 units away from the number 4?” If we draw the number line we see that there are two possibilities: 9 and -1.


2. Solve the equation \begin{align*}|x+3|=2\end{align*} and interpret the answers.

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*} and \begin{align*}x = -1\end{align*} are the answers.

The equation \begin{align*}|x+3|=2\end{align*} can be re-written as: \begin{align*}|x-(-3)|=2\end{align*}. We can interpret this as “what numbers on the number line are 2 units away from -3?” There are two possibilities: -5 and -1.

Solve Real-World Problems Using Absolute Value Equations

Real-World Application: Packing Coffee

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?

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*} is the weight of a bag in ounces, then the equation that describes this problem is \begin{align*}|x-16|=0.25\end{align*}.

Now we must consider the positive and negative options and solve each equation separately:

\begin{align*}& x-16 = 0.25 \qquad \text{and} \quad x-16 =-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*} and \begin{align*}16 - 15.75 = 0.25 \ ounces\end{align*}. The answers are 0.25 ounces bigger and smaller than 16 ounces respectively.

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.



Example 1

Solve the equation \begin{align*}|2x-7|=6\end{align*} and interpret the answers.

Solve the two equations:

\begin{align*}& 2x-7 = 6 \qquad \qquad \quad 2x-7=-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*} and \begin{align*}x=\frac{1}{2}\end{align*}.

The interpretation of this problem is clearer if the equation \begin{align*}|2x-7|=6\end{align*} is divided by 2 on both sides to get \begin{align*}\frac{1}{2}|2x-7|=3\end{align*}. Because \begin{align*}\frac{1}{2}\end{align*} is nonnegative, we can distribute it over the absolute value sign to get \begin{align*}\left | x-\frac{7}{2} \right |=3\end{align*}. The question then becomes “What numbers on the number line are 3 units away from \begin{align*}\frac{7}{2}\end{align*}?” There are two answers: \begin{align*}\frac{13}{2}\end{align*} and \begin{align*}\frac{1}{2}\end{align*}.


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

  1. \begin{align*}|x-5|=10\end{align*}
  2. \begin{align*}|x+2|=6\end{align*}
  3. \begin{align*}|5x-2|=3\end{align*}
  4. \begin{align*}|x-4|=-3\end{align*}
  5. \begin{align*}\left|2x-\frac{1}{2}\right|=10\end{align*}
  6. \begin{align*}|-x+5|=\frac{1}{5}\end{align*}
  7. \begin{align*}\left|\frac{1}{2}x-5\right|=100\end{align*}
  8. \begin{align*}|10x-5|=15\end{align*}
  9. \begin{align*}|0.1x+3|=0.015\end{align*}
  10. \begin{align*}|27-2x|=3x+2\end{align*}

Review (Answers)

To view the Review answers, open this PDF file and look for section 6.8. 

Notes/Highlights Having trouble? Report an issue.

Color Highlighted Text Notes
Please to create your own Highlights / Notes
Show More


Absolute Value

The absolute value of a number is the distance the number is from zero. Absolute values are never negative.

linear equation

A linear equation is an equation between two variables that produces a straight line when graphed.

Image Attributions

Show Hide Details
Difficulty Level:
At Grade
Date Created:
Aug 13, 2012
Last Modified:
Apr 11, 2016
Save or share your relevant files like activites, homework and worksheet.
To add resources, you must be the owner of the Modality. Click Customize to make your own copy.
Please wait...
Please wait...
Image Detail
Sizes: Medium | Original