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Absolute Value Equations

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Solving Absolute Value Equations

To determine the height of skeletal remains, archaeologists use the equation H=2.26f + 66.4 , 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  \pm 3.42 cm . 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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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, |x| .

|5| = 5 && |-9| = 9	&& |0| = 0 && |-1| = 1

When solving an absolute value equation, the value of x 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 |x|=1 , then x can be 1 or -1 because |1| = 1 and |-1| = 1 .

If |x|=15 , then x can be 15 or -15 because |15| = 15 and |-15| = 15 .

From these statements we can conclude:

Example A

Determine if x = -12 is a solution to |2x-5|=29 .

Solution: Plug in -12 for x to see if it works.

|2(-12)-5| &= 29\\|-24-5| &= 29 \\|-29| &= 29

-12 is a solution to this absolute value equation.

Example B

Solve |x+4|=11 .

Solution: There are going to be two answers for this equation. x + 4 can equal 11 or -11.

& \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

Test the solutions:

& |7+4|=11 \qquad |-15+4|=11 \\& \quad \ |11|=11 \  \qquad \ |-11|=11

Example C

Solve \bigg | \frac{2}{3}x-5 \bigg |=17 .

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

& \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

Test the solutions:

& \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

Intro Problem Revisit First, we need to find the height of the skeleton using the equation H=2.26f+66.4 , where f=46.8 .

H&=2.26(46.8)+66.4 \\H&=172.168 cm

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

& \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

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 x = -5 a solution to |3x+22|=6 ?

Solve the following absolute value equations.

2. |6x-11|+2=41

3. \bigg | \frac{1}{2}x+3 \bigg |=9

Answers

1. Plug in -5 for x to see if it works.

|3(-5)+22|=6\\|-15+22|=6\\|-7| \ne 6

-5 is not a solution because |-7| = 7 , 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.

& \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}

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

\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

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

& \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

Test solutions:

& \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

Vocabulary

Absolute Value
The positive distance from zero a given number is.

Practice

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

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

Solve the following absolute value equations.

  1. |x+3|=8
  2. |2x|=9
  3. |2x+15|=3
  4. \bigg |\frac{1}{3}x-5 \bigg |=2
  5. \bigg |\frac{x}{6}+4 \bigg |=5
  6. |7x-12|=23
  7. \bigg |\frac{3}{5}x+2 \bigg |=11
  8. |4x -15|+1=18
  9. |-3x+20|=35
  10. |12x-18|=0
  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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