Skip to main content
K12 LibreTexts

2.4.9: Absolute Value Equations

  1. Last updated
  2. Save as PDF
  • Page ID
    4362

    • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    \( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)

    ( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)

    \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

    \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)

    \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

    \( \newcommand{\Span}{\mathrm{span}}\)

    \( \newcommand{\id}{\mathrm{id}}\)

    \( \newcommand{\Span}{\mathrm{span}}\)

    \( \newcommand{\kernel}{\mathrm{null}\,}\)

    \( \newcommand{\range}{\mathrm{range}\,}\)

    \( \newcommand{\RealPart}{\mathrm{Re}}\)

    \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

    \( \newcommand{\Argument}{\mathrm{Arg}}\)

    \( \newcommand{\norm}[1]{\| #1 \|}\)

    \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

    \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    \( \newcommand{\vectorA}[1]{\vec{#1}}      % arrow\)

    \( \newcommand{\vectorAt}[1]{\vec{\text{#1}}}      % arrow\)

    \( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vectorC}[1]{\textbf{#1}} \)

    \( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)

    \( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)

    \( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)

    \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    \(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)

    Absolute Value Equations

    Suppose that a movie director has been told by a film studio that the length of the movie he's making must not differ from 120 minutes by more than 10 minutes. What would be the longest and shortest acceptable movies he could make? What absolute value equation could you set up to find out?

    Absolute Value Equations

    Absolute value situations can also involve unknown variables. For example, suppose the distance from zero is 16. What two points can this represent?

    Begin by writing an absolute value sentence to represent this situation.

    16=|n|, where n=the missing value

    Which two numbers are 16 units from zero?

    n=16 or n=−16

    Absolute value situations can also involve distances from points other than zero. We treat such cases as compound inequalities, separating the two independent equations and solving separately.

    Let's solve the following absolute value problems:

    1. Solve for x:|x−4|=5.

    This equation looks like the distance definition:

    distance=|x−y| or |y−x|

    The distance is 5, and the value of y is 4. We are looking for two values that are five units away from four on a number line.

    Visually, we can see the answers are –1 and 9.

    Algebraically, we separate the two absolute value equations and solve.

    x−4=5 and x−4=−(5)

    By solving each, the solutions become:

    x=9 and x=−1

    1. Solve |2x−7|=6.

    Begin by separating this into its separate equations.

    2x−7=6 and 2x−7=−6

    Solve each equation independently.

    2x−7=6

    2x−7+7=6+7

    2x=13

    x=13/2

    2x−7=−6

    2x−7+7=−6+7

    2x=1

    x=1/2

    Now, let's apply absolute value equations to a real-world situation and solve the following problem:

    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 and if it is more than 0.25 ounces overweight or underweight, it is emptied and repacked. What are the lightest and heaviest acceptable bags?

    Figure \(\PageIndex{1}\)

    The varying quantity is the weight of the bag of coffee beans. Choosing a letter to represent this quantity and writing an absolute value equation yields:

    |w−16|=0.25

    This equation says that the distance away from 16 is equal to 0.25.

    Separate and solve.

    w−16=0.25

    w=16.25

    w−16=−0.25

    w=15.75

    The lightest bag acceptable is 15.75 ounces and the heaviest bag accepted is 16.25 ounces.

    Examples

    Example 2.4.9.1

    Earlier, you were told that the length of a film by a certain movie director cannot differ from 120 minutes by more than 10 minutes. What would be the longest and shortest acceptable movies he could make? What absolute value equation could you set up to find out?

    Solution

    The quantity that varies is the length of the movie. Let's use m to represent the length of the movie. Putting the information together, the equation to solve is:

    |m−120|=10

    This equation says that the distance away from 120 is 10.

    Separate and solve the equation.

    m−120=10

    m=130

    m−120=−10

    m=110

    The shortest the film can be is 110 minutes and the longest the film can be is 130 minutes.

    Example 2.4.9.2

    Solve for n: |−3n+5|=17.

    Solution

    |−3n+5|=17 means that −3n+5=17 or −3n+5=−17.

    Solving each one separately, we get:

    −3n+5=17

    −3n=12

    n=−4

    −3n+5=−17

    −3n=−22

    n=22/3

    Review

    In 1–12, solve the absolute value equations and interpret the results by graphing the solutions on a number line.

    1. |7u|=77
    2. |x−5|=10
    3. |5r−6|=9
    4. 1=(|6+5z|)/5
    5. |8x|=32
    6. |m/8|=1
    7. |x+2|=6
    8. |5x−2|=3
    9. 51=|1−5b|
    10. 8=3+|10y+5|
    11. |4x−1|=19
    12. 8|x+6|=−48
    1. A company manufactures rulers. Their 12-inch rulers pass quality control if they're within 1/32 inches of the ideal length. What is the longest and shortest ruler that can leave the factory?

    Mixed Review

    1. A map has a scale of 2 inch=125 miles. How far apart would two cities be on the map if the actual distance is 945 miles?
    2. Determine the domain and range: {(−9,0),(−6,0),(−4,0),(0,0),(3,0),(5,0)}.
    3. Is the relation in question #15 a function? Explain your reasoning.
    4. Consider the problem 3(2x−7)=100. Lei says the first step to solving this equation is to use the Distributive Property to cancel the parentheses. Hough says the first step to solving this equation is to divide by 3. Who is right? Explain your answer.
    5. Graph 4x+y=6 using its intercepts.
    6. Write 3/30 as a percent. Round to the nearest hundredth.
    7. Simplify −5(2/3)÷7(1/8).

    Review (Answers)

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

    Vocabulary

    Term Definition
    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.

    Additional Resources

    Video:

    Activity: Absolute Value Equations Discussion Questions

    Practice: Absolute Value Equations

    Real World Application: Calibration

    This page titled 2.4.9: Absolute Value Equations is shared under a CC BY-NC license and was authored, remixed, and/or curated by CK-12 Foundation via source content that was edited to the style and standards of the LibreTexts platform.