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# Absolute Value Equations

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Practice 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.

### Watch This

Watch the first part of this video.

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

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:

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

### Explore More

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 Language: English

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.