ORCCA Absolute Value Equations

Section 13.4 Absolute Value Equations

Objectives: PCC Course Content and Outcome Guide

[cross-reference to target(s) "ccog-solve-absolute-value-equations" missing or not unique]

Whether it’s a washer, nut, bolt, or gear, when a machine part is made, it must be made to fit with all of the other parts of the system. Since no manufacturing process is perfect, there are small deviations from the norm when each piece is made. In fact, manufacturers have a range of acceptable values for each measurement of every screw, bolt, etc.

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Figure 13.4.1. Alternative Video Lesson

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Let’s say we were examining some new bolts just out of the factory. The manufacturer specifies that each bolt should be within a tolerance of 0.04 mm to 10 mm in diameter. So the lowest diameter that the bolt could be to make it through quality assurance is 0.04 mm smaller than 10 mm, which is 9.96 mm. Similarly, the largest diameter that the bolt could be is 0.04 mm larger than 10 mm, which is 10.04 mm.

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To write an equation that describes the minimum and maximum deviation from average, we want the difference between the actual diameter and the specification to be equal to 0.04 mm. Since absolute values are used to describe distances, we can summarize our thoughts mathematically as

| x - 10 | = 0.04,

where

x

represents the diameter of an acceptably sized bolt, in millimeters. This equation says the same thing as the lowest diameter that the bolt could be to make it through quality assurance is 9.96 mm and the largest diameter that the bolt could be is 10.04 mm.

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In this section we will examine a variety of problems that relate to this sort of math with absolute values.

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Subsection 13.4.1 Graphs of Absolute Value Functions

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Absolute value functions have generally the same shape. They are usually described as “V”-shaped graphs and the tip of the “V” is called the vertex. A few graphs of various absolute value functions are shown in Figure 2. In general, the domain of an absolute value function (where there is a polynomial inside the absolute value) is

(- \infty, \infty) .

A Cartesian graph with the graph of y=abs(x), which looks like a V with the vertex of the V at the origin. The right side extends as a line with slope 1 and the left side extends with slope -1. — (a) $y = | x |$

A Cartesian graph with the graph of y=-abs(x+2), which looks like an upside down V with the vertex of the V at the point (-2,0). The right side extends as a line with slope -1 and the left side extends with slope 1. — (a) $y = | x |$

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Example 13.4.3.

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Let

h (x) = - 2 | x - 3 | + 5 .

Using technology, create table of values with

x

-values from

- 3

3,

using an increment of

1 .

Then sketch a graph of

y = h (x) .

State the domain and range of

h .

Explanation.

$x$	$y$
$- 3$	$- 7$
$- 2$	$- 5$
$- 1$	$- 3$
$0$	$- 1$
$1$	$1$
$2$	$3$
$3$	$5$

Figure 13.4.4. Table for

y = h (x) .

A Cartesian graph with the graph of y=-2*abs(x-3)+5, which looks like an upside down V with the vertex of the V at the point (3,5). The right side extends as a line with slope -2 and the left side extends with slope 2. — Figure 13.4.5. Graph of $y = h (x)$

The graph indicates that the domain is

(\infty, \infty)

as it goes to the right and left indefinitely. The range is

(- \infty, 5] .

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Example 13.4.6.

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Let

j (x) = | | x + 1 | - 2 | - 1 .

Using technology, create table of values with

x

-values from

- 5

5,

using an increment of

1

and sketch a graph of

y = j (x) .

State the domain and range of

j .

Explanation.

This is a strange one because it has an absolute value within an absolute value.

$x$	$y$
$- 5$	$1$
$- 4$	$0$
$- 3$	$- 1$
$- 2$	$0$
$- 1$	$1$
$0$	$0$
$1$	$- 1$
$2$	$0$
$3$	$1$
$4$	$2$
$5$	$3$

Figure 13.4.7. A table of values for

y = j (x) .

A Cartesian graph with the graph of y=abs(abs(x+1)-2)+1, which looks like a W with the vertices of the W at the points (-3,-1), (-1,1), and (1,-1). The right side extends as a line with slope 1 and the left side extends with slope -1. — Figure 13.4.8. $y = | | x + 1 | - 2 | - 1$

The graph indicates that the domain is

(\infty, \infty)

as it goes to the right and left indefinitely. The range is

[- 1, \infty) .

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Subsection 13.4.2 Solving Absolute Value Equations with One Absolute Value

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We can solve absolute value equations graphically.

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Example 13.4.9.

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Solve the equations graphically using the graphs provided.

$| x | = 3$

$A Cartesian graph with the graph of y=abs(x);the graph has a V shape with the vertex of the V at the origin;the line on the right side has a slope of 1;the line on the left side has a slope of -1$
$| 2 x + 3 | = 5$

$A Cartesian graph with the graph of y=abs(x);the vertex of the V shape is at the point (-1.5,0);the line on the right side has a slope of 2 and the left side extends with slope -2.$

Explanation.

To solve the equations graphically, first we need to graph the right sides of the equations also.

$| x | = 3$

$A Cartesian graph with the graphs of y=abs(x) and y=3;the vertex of the V shape is at the origin. The right side has a slope of 1 and the left side has a slope of -1;the graph of y=3 is a horizontal line at y=3;the two graphs intersect at (-3,3) and (3,3)$

Since the graph of $y = | x |$ crosses $y = 3$ at the $x$ -values $- 3$ and $3,$ the solution set to the equation $| x | = 3$ must be ${- 3, 3} .$
$| 2 x + 3 | = 5$

$A Cartesian graph with the graphs of y=abs(2x+3) and y=5;the vertex of the V shape is at the point (-1.5,0);the right side has a slope of 2 and the left side has a slope of -2;the graph of y=5 is a horizontal line at y=5;the two graphs intersect at (-4,5) and (1,5)$

Since the graph of $y = | 2 x + 3 |$ crosses $y = 5$ at the $x$ -values $- 4$ and $1,$ the solution set to the equation $| 2 x + 3 | = 5$ must be ${- 4, 1} .$

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Remark 13.4.10.

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Please note that there is a big difference between the expression

| 3 |

and the equation

| x | = 3 .

The expression $| 3 |$ is describing the distance from $0$ to the number $3 .$ The distance is just $3 .$ So $| 3 | = 3 .$
The equation $| x | = 3$ is asking you to find the numbers that are a distance of $3$ from $0 .$ These two numbers are $3$ and $- 3 .$

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Let’s solve some absolute value equations algebraically. To motivate this, we will think about what an absolute value equation means in terms of the “distance from zero” definition of absolute value. If

| X | = n,

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where

n \geq 0,

then this means that we want all of the numbers,

X,

that are a distance

n

from

0 .

Since we can only go left or right along the number line, this is describing both

X = n

as well as

X = - n .

a number line graph with a point labeled at n and -n;both points are n units from 0 — 🔗
Figure 13.4.11. A Numberline with Points a Distance $n$ from $0$

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Let’s summarize this.

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Fact 13.4.12. Equations with an Absolute Value Expression.

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Let

n

be a non-negative number and

X

be an algebraic expression. Then the equation

| X | = n

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has the same solutions as

X = n or X = - n .

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Example 13.4.13.

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Solve the absolute value equations using Fact 12. Write solutions in a solution set.

$| x | = 6$
$| x | = - 4$
$| 5 x - 7 | = 23$
$| 14 - 3 x | = 8$
$| 3 - 4 x | = 0$

Explanation.

Fact 12 says that the equation $| x | = 6$ is the same as

$x = 6 or x = - 6 .$

Thus the solution set is ${6, - 6} .$
Fact 12 doesn’t actually apply to the equation $| x | = - 4$ because the value on the right side is negative. How often is an absolute value of a number negative? Never! Thus, there are no solutions and the solution set is the empty set, denoted $\emptyset .$
The equation $| 5 x - 7 | = 23$ breaks into two pieces, each of which needs to be solved independently.

$\begin{aligned} 5 x - 7 & = 23 & or & 5 x - 7 & = - 23 \\ 5 x & = 30 & or & 5 x & = - 16 \\ x & = 6 & or & x & = - \frac{16}{5} \end{aligned}$

Thus the solution set is ${6, - \frac{16}{5}} .$
The equation $| 14 - 3 x | = 8$ breaks into two pieces, each of which needs to be solved independently.

$\begin{aligned} 14 - 3 x & = 8 & or & 14 - 3 x & = - 8 \\ - 3 x & = - 6 & or & - 3 x & = - 22 \\ x & = 2 & or & x & = \frac{22}{3} \end{aligned}$

Thus the solution set is ${2, \frac{22}{3}} .$
The equation $| 3 - 4 x | = 0$ breaks into two pieces, each of which needs to be solved independently.

$\begin{aligned} 3 - 4 x & = 0 & or & 3 - 4 x & = - 0 \end{aligned}$
Since these are identical equations, all we have to do is solve one equation.
$\begin{aligned} 3 - 4 x & = 0 \\ - 4 x & = - 3 \\ x & = \frac{3}{4} \end{aligned}$

Thus, the equation $| 3 - 4 x | = 0$ only has one solution, and the solution set is ${\frac{3}{4}} .$

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Subsection 13.4.3 Solving Absolute Value Equations with Two Absolute Values

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Example 13.4.14.

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Let’s graphically solve an equation with an absolute value expression on each side:

| x | = | 2 x + 6 | .

Since

| x | = 3

had two solutions as we saw in Example 9, you might be wondering how many solutions

| x | = | 2 x + 6 |

will have. Make a graph to find out what the solutions of the equation are.

Explanation.

a Cartesian graph with the graphs of y=abs(x) and y=abs(2x+6);y=abs(x) has its vertex at (0,0) with a slope of 1 on the right and -1 on the left;y=abs(2x+6) has its vertex at (-3,0) with a slope of 2 on the right and -2 on the left;the V-shaped graphs cross at the points (-2,2) and (-6,6) — Figure 13.4.15. $y = | x |$ and $y = | 2 x + 6 |$

Figure 15 shows that there are also two points of intersection between the graphs of

y = | x |

and

y = | 2 x + 6 | .

The solutions to the equation

| x | = | 2 x + 6 |

are the

x

-values where the graphs cross. So, the solution set is

{- 6, - 2} .

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Example 13.4.16.

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Solve the equation

| x + 1 | = | 2 x - 4 |

graphically.

Explanation.

First break up the equation into the left side and the right side and graph each separately, as in

y = | x + 1 |

and

y = | 2 x - 4 | .

We can see in the graph that the graphs intersect twice. The

x

-values of those intersections are

1

and

5

so the solution set to the equation

| x + 1 | = | 2 x - 4 |

{1, 5} .

a Cartesian graph with the graphs of y=abs(x+1) and y=abs(2x-4);y=abs(x+1) has its vertex at (-1,0) with a slope of 1 on the right and -1 on the left;y=abs(2x-4) has its vertex at (2,0) with a slope of 2 on the right and -2 on the left;the V-shaped graphs cross at the points (1,2) and (5,6) — Figure 13.4.17. $y = | x + 1 |$ and $y = | 2 x - 4 |$

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Fortunately, this kind of equation also has a rule to solve these types of equations algebraically that is similar to the rule for equations with one absolute value.

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Fact 13.4.18. Equations with Two Absolute Value Expressions.

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Let

X

and

Y

be linear algebraic expressions. Then, the equation

| X | = | Y |

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has the same solutions as

X = Y or X = - Y .

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Remark 13.4.19.

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You might wonder why the negative sign “has” to go on the right side of the equation in

X = - Y .

It doesn’t; it can go on either side of the equation. The equations

X = - Y

and

- X = Y

are equivalent. Similarly,

- X = - Y

is equivalent to

X = Y .

That’s why we only need to solve two of the four possible equations.

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Example 13.4.20.

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Solve the equations using Fact 18.

$| x - 4 | = | 3 x - 2 |$
$| \frac{1}{2} x + 1 | = | \frac{1}{3} x + 2 |$
$| x - 2 | = | x + 1 |$
$| x - 1 | = | 1 - x |$

Explanation.

The equation $| x - 4 | = | 3 x - 2 |$ breaks down into two pieces:

$\begin{aligned} x - 4 & = 3 x - 2 & or & x - 4 & = - (3 x - 2) \\ x - 4 & = 3 x - 2 & or & x - 4 & = - 3 x + 2 \\ - 2 & = 2 x & or & 4 x & = 6 \\ \frac{- 2}{2} & = \frac{2 x}{2} & or & \frac{4 x}{4} & = \frac{6}{4} \\ - 1 & = x & or & x & = \frac{3}{2} \end{aligned}$

So, the solution set is ${- 1, \frac{3}{2}} .$
The equation $| \frac{1}{2} x + 1 | = | \frac{1}{3} x + 2 |$ breaks down into two pieces:

$\begin{aligned} \frac{1}{2} x + 1 & = \frac{1}{3} x + 2 & or & \frac{1}{2} x + 1 & = - (\frac{1}{3} x + 2) \\ \frac{1}{2} x + 1 & = \frac{1}{3} x + 2 & or & \frac{1}{2} x + 1 & = - \frac{1}{3} x - 2 \\ 6 \cdot (\frac{1}{2} x + 1) & = 6 \cdot (\frac{1}{3} x + 2) & or & 6 \cdot (\frac{1}{2} x + 1) & = 6 \cdot (- \frac{1}{3} x - 2) \\ 3 x + 6 & = 2 x + 12 & or & 3 x + 6 & = - 2 x - 12 \\ x & = 6 & or & 5 x & = - 18 \\ x & = 6 & or & x & = - \frac{18}{5} \end{aligned}$

So, the solution set is ${6, - \frac{18}{5}} .$
The equation $| x - 2 | = | x + 1 |$ breaks down into two pieces:

$\begin{aligned} x - 2 & = x + 1 & or & x - 2 & = - (x + 1) \\ x - 2 & = x + 1 & or & x - 2 & = - x - 1 \\ x & = x + 3 & or & 2 x & = 1 \\ 0 & = 3 & or & x & = \frac{1}{2} \end{aligned}$

Note that one of the two pieces gives us an equation with no solutions. Since $0 \neq 3,$ we can safely ignore this piece. Thus the only solution is $\frac{1}{2} .$

We should visualize this equation graphically because our previous assumption was that two absolute value graphs would cross twice. The graph shows why there is only one crossing: the left and right sides of each “V” are parallel.

$a Cartesian graph with the graphs of y=abs(x-2) and y=abs(x+1);y=abs(x-2) has its vertex at (2,0) with a slope of 1 on the right and -1 on the left;y=abs(x+1) has its vertex at (-1,0) with a slope of 1 on the right and -1 on the left;the V-shaped graphs cross at only one point (1/2,3/2)$
The equation $| x - 1 | = | 1 - x |$ breaks down into two pieces:

$\begin{aligned} x - 1 & = 1 - x & or & x - 1 & = - (1 - x) \\ x - 1 & = 1 - x & or & x - 1 & = - 1 + x \\ 2 x & = 2 & or & x & = 0 + x \\ x & = 1 & or & 0 & = 0 \end{aligned}$

Note that our second equation is an identity so recall from Section 2.4 that the solution set is “all real numbers.”

So, our two pieces have solutions $1$ and “all real numbers.” Since $1$ is a real number and we have an or statement, our overall solution set is $(- \infty, \infty) .$ The graph confirms our answer since the two “V” graphs are coinciding.

$a Cartesian graph with the graphs of y=abs(x-1) and y=abs(1-x);y=abs(x-1) has its vertex at (1,0) with a slope of 1 on the right and -1 on the left;y=abs(1-x) has its vertex at (1,0) with a slope of 1 on the right and -1 on the left;the V-shaped graphs overlap everywhere so they are coinciding$

Figure 13.4.21. $y = | x - 1 |$ and $y = | 1 - x |$

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Solving an absolute value equation like

| 2 x + 1 | = 3

is “easy” because we can turn it into two equations of what simpler type?

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7.

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Write the equation $5 = | 4 x | - 3$ as two separate equations. Neither of your equations should use absolute value.

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Solve the original equation by solveing each of the equations above.

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8.

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Write the equation $5 = | 7 x | - 6$ as two separate equations. Neither of your equations should use absolute value.

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Solve the original equation by solveing each of the equations above.

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9.

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Write the equation $| 6 - \frac{r}{5} | = 3$ as two separate equations. Neither of your equations should use absolute value.
Solve the equation $| 6 - \frac{r}{5} | = 3,$ possibly by solving each equation from part a.

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10.

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Write the equation $| 8 - \frac{r}{3} | = 5$ as two separate equations. Neither of your equations should use absolute value.
Solve the equation $| 8 - \frac{r}{3} | = 5,$ possibly by solving each equation from part a.

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11.

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Verify that the value $- 1$ is a solution to the absolute value equation $| \frac{x - 3}{2} | = 2 .$
Verify that the value $\frac{2}{3}$ is a solution to the absolute value inequality $| 6 x - 5 | < 4 .$