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Absolute Value Equations

Equations may have two solutions due to the absolute value operation

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Solving Absolute Value Equations

To determine the height of skeletal remains, archaeologists use the equation \begin{align*}H=2.26f + 66.4\end{align*}, where H is the height in centimeters and f is the length of the skeleton's femur (also in cm). The equation has a margin of error of \begin{align*} \pm 3.42 cm\end{align*}. Dr. Jordan found a skeletal femur that is 46.8 cm. Determine the greatest height and the least height of this person.

Absolute Value Equations

Absolute value is the distance a number is from zero. Because distance is always positive, the absolute value will always be positive. Absolute value is denoted with two vertical lines around a number, \begin{align*}|x|\end{align*}.

\begin{align*}|5| = 5 && |-9| = 9 && |0| = 0 && |-1| = 1\end{align*}

When solving an absolute value equation, the value of \begin{align*}x\end{align*} could be two different possibilities; whatever makes the absolute value positive OR whatever makes it negative. Therefore, there will always be TWO answers for an absolute value equation.

If \begin{align*}|x|=1\end{align*}, then \begin{align*}x\end{align*} can be 1 or -1 because \begin{align*}|1| = 1\end{align*} and \begin{align*}|-1| = 1\end{align*}.

If \begin{align*}|x|=15\end{align*}, then \begin{align*}x\end{align*} can be 15 or -15 because \begin{align*}|15| = 15\end{align*} and \begin{align*}|-15| = 15\end{align*}.

From these statements we can conclude:

 

Let's determine if \begin{align*}x = -12\end{align*} is a solution to \begin{align*}|2x-5|=29\end{align*}.

Plug in -12 for \begin{align*}x\end{align*} to see if it works.

\begin{align*}|2(-12)-5| &= 29\\ |-24-5| &= 29 \\ |-29| &= 29\end{align*}

-12 is a solution to this absolute value equation.

Now, let's solve the following absolute value equations.

  1. Solve \begin{align*}|x+4|=11\end{align*}.

There are going to be two answers for this equation. \begin{align*}x + 4\end{align*} can equal 11 or -11.

\begin{align*}& \qquad \quad |x+4|=11\\ & \qquad \quad \ \ \swarrow \searrow\\ & x+4 = 11 \quad x+4=-11\\ & \qquad \qquad \ \ or \qquad \qquad\\ & \quad \ \ x =7 \qquad \quad x=-15\end{align*}

Test the solutions:

\begin{align*}& |7+4|=11 \qquad |-15+4|=11 \\ & \quad \ |11|=11 \ \qquad \ |-11|=11 \end{align*}

  1. Solve \begin{align*}\bigg | \frac{2}{3}x-5 \bigg |=17\end{align*}.

Here, what is inside the absolute value can be equal to 17 or -17.

\begin{align*}& \qquad \quad \qquad \bigg | \frac{2}{3}x-5 \bigg |=17\\ & \qquad \qquad \quad \ \swarrow \searrow\\ & \frac{2}{3}x-5=17 \qquad \quad \frac{2}{3}x-5=-17\\ & \quad \ \ \frac{2}{3}x=22 \qquad or \quad \ \ \frac{2}{3}x=-12\\ & \qquad \ x=22 \cdot \frac{3}{2} \qquad \qquad x=-12 \cdot \frac{3}{2}\\ & \qquad \ x=33 \qquad \quad \quad \quad \ x=-18\end{align*}

Test the solutions:

\begin{align*}& \bigg | \frac{2}{3}(33)-5 \bigg |=17 \qquad \quad \ \bigg | \frac{2}{3}(-18)-5\bigg |=17\\ & \quad \ \ |22-5|=17 \ \qquad \ \ |-12-5|=17\\ & \qquad \quad \ |17|=17 \qquad \qquad \qquad |-17|=17\end{align*}

Examples

Example 1

Earlier, you were asked to determine the greatest height and the least height of the skeletons. 

First, we need to find the height of the skeleton using the equation \begin{align*}H=2.26f+66.4\end{align*}, where \begin{align*}f=46.8\end{align*}.

\begin{align*}H&=2.26(46.8)+66.4 \\ H&=172.168 cm\end{align*}

Now, let's use an absolute value equation to determine the margin of error, and thus the greatest and least heights.

\begin{align*}& \qquad \qquad \qquad \quad |x-172.168|=3.42\\ & \qquad \qquad \qquad \quad \swarrow \searrow\\ & x-172.168 = 3.42 \quad x-172.168=-3.42\\ & \qquad \qquad \qquad \qquad or \qquad \qquad\\ & \quad \ \ x =175.588 \qquad \quad x=168.748\end{align*}

So the person could have been a maximum of 175.588 cm or a minimum of 168.748 cm. In inches, this would be 69.13 and 66.44 inches, respectively.

Example 2

Is \begin{align*}x = -5\end{align*} a solution to \begin{align*}|3x+22|=6\end{align*}?

Plug in -5 for \begin{align*}x\end{align*} to see if it works.

\begin{align*}|3(-5)+22|=6\\ |-15+22|=6\\ |-7| \ne 6\end{align*}

-5 is not a solution because \begin{align*}|-7| = 7\end{align*}, not 6.

Example 3

Solve the following absolute value equation: \begin{align*}|6x-11|+2=41\end{align*}.

Find the two solutions. Because there is a 2 being added to the left-side of the equation, we first need to subtract it from both sides so the absolute value is by itself.

\begin{align*}& \qquad \ |6x-11|+2=41\\ & \qquad \qquad |6x-11|=39\\ & \qquad \quad \qquad \swarrow \searrow\\ & \ 6x-11=39 \quad 6x-11=-39\\ & \qquad \ 6x=50 \qquad \quad \ 6x=-28\\ & \qquad \quad x=\frac{50}{6} \quad or \quad \ \ x=-\frac{28}{6}\\ & \quad \qquad \ = \frac{25}{3} \ or \ 8 \frac{1}{3} \quad \ = - \frac{14}{3} \ or \ -4 \frac{2}{3}\end{align*}

Check both solutions. It is easier to check solutions when they are improper fractions.

\begin{align*}\bigg |6 \left(\frac{25}{3}\right) -11 \bigg| & = 39 \qquad \qquad \quad \bigg |6 \left(- \frac{14}{3} \right) -11 \bigg| = 39\\ |50-11| &= 39 \ \quad and \qquad \ |-28-11| = 39 \\ |39| &= 39 \qquad \qquad \qquad \qquad \ \ |-39| = 39\end{align*}

Example 4

Solve the following absolute value equation: \begin{align*}\bigg | \frac{1}{2}x+3 \bigg |=9\end{align*}.

What is inside the absolute value is equal to 9 or -9.

\begin{align*}& \qquad \bigg | \frac{1}{2}x+3\bigg |=9\\ & \qquad \quad \swarrow \searrow\\ & \frac{1}{2}x+3=9 \quad \frac{1}{2}x+3=-9\\ & \quad \ \ \frac{1}{2}x=6 \quad or \quad \frac{1}{2}x=-12\\ & \qquad \ x=12 \qquad \quad x=-24\end{align*}

Test solutions:

\begin{align*}& \bigg | \frac{1}{2} (12) +3 \bigg |=9 \qquad \quad \bigg | \frac{1}{2} (-24) +3 \bigg |=9\\ & \qquad |6+3|=9 \ \qquad \ \ |-12+3|=9 \\ & \qquad \quad \ \ |9|=9 \qquad \qquad \qquad \ \ |-9|=9\end{align*}

Review

Determine if the following numbers are solutions to the given absolute value equations.

  1. \begin{align*}|x-7|=16;9\end{align*}
  2. \begin{align*}|\frac{1}{4}x+1|=4;-8\end{align*}
  3. \begin{align*}|5x-2|=7;-1\end{align*}

Solve the following absolute value equations.

  1. \begin{align*}|x+3|=8\end{align*}
  2. \begin{align*}|2x|=9\end{align*}
  3. \begin{align*}|2x+15|=3\end{align*}
  4. \begin{align*}\bigg |\frac{1}{3}x-5 \bigg |=2\end{align*}
  5. \begin{align*}\bigg |\frac{x}{6}+4 \bigg |=5\end{align*}
  6. \begin{align*}|7x-12|=23\end{align*}
  7. \begin{align*}\bigg |\frac{3}{5}x+2 \bigg |=11\end{align*}
  8. \begin{align*}|4x -15|+1=18\end{align*}
  9. \begin{align*}|-3x+20|=35\end{align*}
  10. \begin{align*}|12x-18|=0\end{align*}
  11. What happened in #13? Why do you think that is?
  12. Challenge When would an absolute value equation have no solution? Give an example.

Answers for Review Problems

To see the Review answers, open this PDF file and look for section 1.13. 

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Vocabulary

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.

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