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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.

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Watch the first part of this video.

Khan Academy: Absolute Value Equations

Guidance

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:

Example A

Determine if @$\begin{align*}x = -12\end{align*}@$ is a solution to @$\begin{align*}|2x-5|=29\end{align*}@$ .

Solution: 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.

Example B

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

Solution: 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:

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

Example C

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

Solution: 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:

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

Intro Problem Revisit 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.

Guided Practice

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

Solve the following absolute value equations.

2. @$\begin{align*}|6x-11|+2=41\end{align*}@$

3. @$\begin{align*}\bigg | \frac{1}{2}x+3 \bigg |=9\end{align*}@$

Answers

1. 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.

2. 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.

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

3. 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:

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

Explore More

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.

Vocabulary

Absolute Value

Absolute Value

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

linear equation

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

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