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# 6.4: Absolute Value Equations and Inequalities

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

• Solve an absolute value equation.
• Analyze solutions to absolute value equations.
• Graph absolute value functions.
• Solve absolute value inequalities.
• Rewrite and solve absolute value inequalities as compound inequalities.
• Solve real-world problems using absolute value equations and inequalities.

## Introduction

Timmy is trying out his new roller skates. He’s not allowed to cross the street yet, so he skates back and forth in front of his house. If he skates 20 yards east and then 10 yards west, how far is he from where he started? What if he skates 20 yards west and then 10 yards east?

The absolute value of a number is its distance from zero on a number line. There are always two numbers on the number line that are the same distance from zero. For instance, the numbers 4 and -4 are each a distance of 4 units away from zero.

\begin{align*}|4|\end{align*} represents the distance from 4 to zero, which equals 4.

\begin{align*}|-4|\end{align*} represents the distance from -4 to zero, which also equals 4.

In fact, for any real number \begin{align*}x\end{align*}:

\begin{align*}|x| = x\end{align*} if \begin{align*}x\end{align*} is not negative, and \begin{align*}|x| = -x\end{align*} if \begin{align*}x\end{align*} is negative.

Absolute value has no effect on a positive number, but changes a negative number into its positive inverse.

Example 1

Evaluate the following absolute values.

a) \begin{align*}|25|\end{align*}

b) \begin{align*}|-120|\end{align*}

c) \begin{align*}|-3|\end{align*}

d) \begin{align*}|55|\end{align*}

e) \begin{align*}\left | - \frac{5}{4} \right |\end{align*}

Solution

a) \begin{align*}|25|= 25\end{align*} Since 25 is a positive number, the absolute value does not change it.

b) \begin{align*}|-120| = 120\end{align*} Since -120 is a negative number, the absolute value makes it positive.

c) \begin{align*}|-3| = 3\end{align*} Since -3 is a negative number, the absolute value makes it positive.

d) \begin{align*}|55| = 55\end{align*} Since 55 is a positive number, the absolute value does not change it.

e) \begin{align*}\left | -\frac{5}{4} \right | =\frac{5}{4}\end{align*} Since \begin{align*}-\frac{5}{4}\end{align*} is a negative number, the absolute value makes it positive.

Absolute value is very useful in finding the distance between two points on the number line. The distance between any two points \begin{align*}a\end{align*} and \begin{align*}b\end{align*} on the number line is \begin{align*}|a - b|\end{align*} or \begin{align*}|b - a|\end{align*}.

For example, the distance from 3 to -1 on the number line is \begin{align*}|3 - (-1)| = |4| = 4\end{align*}.

We could have also found the distance by subtracting in the opposite order: \begin{align*}|-1 - 3| = |-4| = 4\end{align*}. This makes sense because the distance is the same whether you are going from 3 to -1 or from -1 to 3.

Example 2

Find the distance between the following points on the number line.

a) 6 and 15

b) -5 and 8

c) -3 and -12

Solution

Distance is the absolute value of the difference between the two points.

a) \begin{align*}\text{distance} = |6 - 15| = |-9| = 9\end{align*}

b) \begin{align*}\text{distance} = |-5 - 8| = |-13| = 13\end{align*}

c) \begin{align*}\text{distance} = |-3 - (-12)| = |9| = 9\end{align*}

Remember: When we computed the change in \begin{align*}x\end{align*} and the change in \begin{align*}y\end{align*} as part of the slope computation, these values were positive or negative, depending on the direction of movement. In this discussion, “distance” means a positive distance only.

## Solve an Absolute Value Equation

We now want to solve equations involving absolute values. Consider the following equation:

\begin{align*}|x|=8\end{align*}

This means that the distance from the number \begin{align*}x\end{align*} to zero is 8. There are two numbers that satisfy this condition: 8 and -8.

When we solve absolute value equations we always consider two possibilities:

1. The expression inside the absolute value sign is not negative.
2. The expression inside the absolute value sign is negative.

Then we solve each equation separately.

Example 3

Solve the following absolute value equations.

a) \begin{align*}|x| = 3\end{align*}

b) \begin{align*}|x| = 10\end{align*}

Solution

a) There are two possibilities: \begin{align*}x = 3\end{align*} and \begin{align*}x = -3\end{align*}.

b) There are two possibilities: \begin{align*}x = 10\end{align*} and \begin{align*}x = -10\end{align*}.

## Analyze Solutions to Absolute Value Equations

Example 4

Solve the equation \begin{align*}|x-4|=5\end{align*} and interpret the answers.

Solution

We consider two possibilities: the expression inside the absolute value sign is nonnegative or is negative. Then we solve each equation separately.

\begin{align*}& x-4 = 5 \quad \text{and} \quad x-4=-5\\ & \quad \ \ x=9 \qquad \qquad \quad \ x=-1\end{align*}

\begin{align*}x = 9\end{align*} and \begin{align*}x = -1\end{align*} are the solutions.

The equation \begin{align*}|x-4|=5\end{align*} can be interpreted as “what numbers on the number line are 5 units away from the number 4?” If we draw the number line we see that there are two possibilities: 9 and -1.

Example 5

Solve the equation \begin{align*}|x+3|=2\end{align*} and interpret the answers.

Solution

Solve the two equations:

\begin{align*}& x+3 = 2 \quad \text{and} \quad \ \ x+3=-2\\ & \quad \ \ x=-1 \qquad \qquad \quad \ x=-5\end{align*}

\begin{align*}x = -5\end{align*} and \begin{align*}x = -1\end{align*} are the answers.

The equation \begin{align*}|x+3|=2\end{align*} can be re-written as: \begin{align*}|x-(-3)|=2\end{align*}. We can interpret this as “what numbers on the number line are 2 units away from -3?” There are two possibilities: -5 and -1.

Example 6

Solve the equation \begin{align*}|2x-7|=6\end{align*} and interpret the answers.

Solution

Solve the two equations:

\begin{align*}& 2x-7 = 6 \qquad \qquad \quad 2x-7=-6\\ & \quad \ \ 2x=13 \qquad \text{and} \qquad \ \ 2x=1\\ & \quad \ \ \ x=\frac{13}{2} \qquad \qquad \qquad \ \ x=\frac{1}{2}\end{align*}

Answer: \begin{align*}x=\frac{13}{2}\end{align*} and \begin{align*}x=\frac{1}{2}\end{align*}.

The interpretation of this problem is clearer if the equation \begin{align*}|2x-7|=6\end{align*} is divided by 2 on both sides to get \begin{align*}\frac{1}{2}|2x-7|=3\end{align*}. Because \begin{align*}\frac{1}{2}\end{align*} is nonnegative, we can distribute it over the absolute value sign to get \begin{align*}\left | x-\frac{7}{2} \right |=3\end{align*}. The question then becomes “What numbers on the number line are 3 units away from \begin{align*}\frac{7}{2}\end{align*}?” There are two answers: \begin{align*}\frac{13}{2}\end{align*} and \begin{align*}\frac{1}{2}\end{align*}.

## Graph Absolute Value Functions

Now let’s look at how to graph absolute value functions.

Consider the function \begin{align*}y=|x-1|\end{align*}. We can graph this function by making a table of values:

\begin{align*}x\end{align*} \begin{align*}y=|x-1|\end{align*}
-2 \begin{align*}y=|-2-1|=|-3|=3\end{align*}
-1 \begin{align*}y=|-1-1|=|-2|=2\end{align*}
0 \begin{align*}y=|0-1|=|-1|=1\end{align*}
1 \begin{align*}y=|1-1|=|0|=0\end{align*}
2 \begin{align*}y=|2-1|=|1|=1\end{align*}
3 \begin{align*}y=|3-1|=|2|=2\end{align*}
4 \begin{align*}y=|4-1|=|3|=3\end{align*}

You can see that the graph of an absolute value function makes a big “V”. It consists of two line rays (or line segments), one with positive slope and one with negative slope, joined at the vertex or cusp.

We’ve already seen that to solve an absolute value equation we need to consider two options:

1. The expression inside the absolute value is not negative.
2. The expression inside the absolute value is negative.

Combining these two options gives us the two parts of the graph.

For instance, in the above example, the expression inside the absolute value sign is \begin{align*}x-1\end{align*}. By definition, this expression is nonnegative when \begin{align*}x-1 \ge 0\end{align*}, which is to say when \begin{align*}x \ge 1\end{align*}. When the expression inside the absolute value sign is nonnegative, we can just drop the absolute value sign. So for all values of \begin{align*}x\end{align*} greater than or equal to 1, the equation is just \begin{align*}y=x-1\end{align*}.

On the other hand, when \begin{align*}x-1 < 0\end{align*} — in other words, when \begin{align*}x < 1\end{align*} — the expression inside the absolute value sign is negative. That means we have to drop the absolute value sign but also multiply the expression by -1. So for all values of \begin{align*}x\end{align*} less than 1, the equation is \begin{align*}y=-(x-1)\end{align*}, or \begin{align*}y=-x+1\end{align*}.

These are both graphs of straight lines, as shown above. They meet at the point where \begin{align*}x-1=0\end{align*} — that is, at \begin{align*}x=1\end{align*}.

We can graph absolute value functions by breaking them down algebraically as we just did, or we can graph them using a table of values. However, when the absolute value equation is linear, the easiest way to graph it is to combine those two techniques, as follows:

1. Find the vertex of the graph by setting the expression inside the absolute value equal to zero and solving for \begin{align*}x\end{align*}.
2. Make a table of values that includes the vertex, a value smaller than the vertex, and a value larger than the vertex. Calculate the corresponding values of \begin{align*}y\end{align*} using the equation of the function.
3. Plot the points and connect them with two straight lines that meet at the vertex.

Example 7

Graph the absolute value function \begin{align*}y=|x+5|\end{align*}.

Solution

Step 1: Find the vertex by solving \begin{align*}x + 5 = 0\end{align*}. The vertex is at \begin{align*}x = -5\end{align*}.

Step 2: Make a table of values:

\begin{align*}x\end{align*} \begin{align*}y=|x+5|\end{align*}
-8 \begin{align*}y=|-8+5|=|-3|=3\end{align*}
-5 \begin{align*}y=|-5+5|=|0|=0\end{align*}
-2 \begin{align*}y=|-2+5|=|3|=3\end{align*}

Step 3: Plot the points and draw two straight lines that meet at the vertex:

Example 8

Graph the absolute value function: \begin{align*}y=|3x-12|\end{align*}

Solution

Step 1: Find the vertex by solving \begin{align*}3x - 12 = 0\end{align*}. The vertex is at \begin{align*}x = 4\end{align*}.

Step 2: Make a table of values:

\begin{align*}x\end{align*} \begin{align*}y=|3x-12|\end{align*}
0 \begin{align*}y=|3(0)-12|=|-12|=12\end{align*}
4 \begin{align*}y=|3(4)-12|=|0|=0\end{align*}
8 \begin{align*}y=|3(8)-12|=|12|=12\end{align*}

Step 3: Plot the points and draw two straight lines that meet at the vertex.

## Solve Real-World Problems Using Absolute Value Equations

Example 9

A company packs coffee beans in airtight bags. Each bag should weigh 16 ounces, but it is hard to fill each bag to the exact weight. After being filled, each bag is weighed; if it is more than 0.25 ounces overweight or underweight, it is emptied and repacked. What are the lightest and heaviest acceptable bags?

Solution

The weight of each bag is allowed to be 0.25 ounces away from 16 ounces; in other words, the difference between the bag’s weight and 16 ounces is allowed to be 0.25 ounces. So if \begin{align*}x\end{align*} is the weight of a bag in ounces, then the equation that describes this problem is \begin{align*}|x-16|=0.25\end{align*}.

Now we must consider the positive and negative options and solve each equation separately:

\begin{align*}& x-16 = 0.25 \qquad \text{and} \quad x-16 =-0.25\\ & \qquad x=16.25 \qquad \qquad \qquad \ \ x=15.75\end{align*}

The lightest acceptable bag weighs 15.75 ounces and the heaviest weighs 16.25 ounces.

We see that \begin{align*}16.25 - 16 = 0.25 \ ounces\end{align*} and \begin{align*}16 - 15.75 = 0.25 \ ounces\end{align*}. The answers are 0.25 ounces bigger and smaller than 16 ounces respectively.

The answer you just found describes the lightest and heaviest acceptable bags of coffee beans. But how do we describe the total possible range of acceptable weights? That’s where inequalities become useful once again.

## Absolute Value Inequalities

Absolute value inequalities are solved in a similar way to absolute value equations. In both cases, you must consider the same two options:

1. The expression inside the absolute value is not negative.
2. The expression inside the absolute value is negative.

Then you must solve each inequality separately.

## Solve Absolute Value Inequalities

Consider the inequality \begin{align*}|x| \le 3\end{align*}. Since the absolute value of \begin{align*}x\end{align*} represents the distance from zero, the solutions to this inequality are those numbers whose distance from zero is less than or equal to 3. The following graph shows this solution:

Notice that this is also the graph for the compound inequality \begin{align*}-3 \le x \le 3\end{align*}.

Now consider the inequality \begin{align*}|x|>2\end{align*}. Since the absolute value of \begin{align*}x\end{align*} represents the distance from zero, the solutions to this inequality are those numbers whose distance from zero are more than 2. The following graph shows this solution.

Notice that this is also the graph for the compound inequality \begin{align*}x < -2\end{align*} or \begin{align*}x >2\end{align*}.

Example 1

Solve the following inequalities and show the solution graph.

a) \begin{align*}|x|<5\end{align*}

b) \begin{align*}|x| \ge 2.5\end{align*}

Solution

a) \begin{align*}|x|<5\end{align*} represents all numbers whose distance from zero is less than 5.

This answer can be written as “\begin{align*}-5 < x < 5\end{align*}”.

b) \begin{align*}|x| \ge 2.5\end{align*} represents all numbers whose distance from zero is more than or equal to 2.5

This answer can be written as “\begin{align*}x \le -2.5\end{align*} or \begin{align*}x \ge 2.5\end{align*}”.

## Rewrite and Solve Absolute Value Inequalities as Compound Inequalities

In the last section you saw that absolute value inequalities are compound inequalities.

Inequalities of the type \begin{align*}|x| can be rewritten as “\begin{align*}-a < x < a\end{align*}”.

Inequalities of the type \begin{align*}|x|>b\end{align*} can be rewritten as “\begin{align*}x < -b\end{align*} or \begin{align*}x >b\end{align*}.”

To solve an absolute value inequality, we separate the expression into two inequalities and solve each of them individually.

Example 2

Solve the inequality \begin{align*}|x-3|<7\end{align*} and show the solution graph.

Solution

Re-write as a compound inequality: \begin{align*}-7

Write as two separate inequalities: \begin{align*}x-3<7\end{align*} and \begin{align*}x-3>-7\end{align*}

Solve each inequality: \begin{align*}x<10\end{align*} and \begin{align*}x > -4\end{align*}

Re-write solution: \begin{align*}-4 < x < 10\end{align*}

The solution graph is

We can think of the question being asked here as “What numbers are within 7 units of 3?”; the answer can then be expressed as “All the numbers between -4 and 10.”

Example 3

Solve the inequality \begin{align*}|4x+5| \le 13\end{align*} and show the solution graph.

Solution

Re-write as a compound inequality: \begin{align*}-13 \le 4x+5 \le 13\end{align*}

Write as two separate inequalities: \begin{align*}4x+5 \le 13\end{align*} and \begin{align*}4x +5 \ge -13\end{align*}

Solve each inequality: \begin{align*}4x \le 8\end{align*} and \begin{align*}4x \ge -18\end{align*}

\begin{align*}x \le 2\end{align*} and \begin{align*}x \ge -\frac{9}{2}\end{align*}

Re-write solution: \begin{align*}-\frac{9}{2} \le x \le 2\end{align*}

The solution graph is

Example 4

Solve the inequality \begin{align*}|x+12|>2\end{align*} and show the solution graph.

Solution

Re-write as a compound inequality: \begin{align*}x+12 < -2\end{align*} or \begin{align*}x+12 > 2\end{align*}

Solve each inequality: \begin{align*}x < -14\end{align*} or \begin{align*}x > -10\end{align*}

The solution graph is

Example 5

Solve the inequality \begin{align*}|8x-15| \ge 9\end{align*} and show the solution graph.

Solution

Re-write as a compound inequality: \begin{align*}8x-15 \le -9\end{align*} or \begin{align*}8x-15 \ge 9\end{align*}

Solve each inequality: \begin{align*}8x \le 6\end{align*} or \begin{align*}8x \ge 24\end{align*}

\begin{align*}x \le \frac{3}{4}\end{align*} or \begin{align*}x \ge 3\end{align*}

The solution graph is

## Solve Real-World Problems Using Absolute Value Inequalities

Absolute value inequalities are useful in problems where we are dealing with a range of values.

Example 6

The velocity of an object is given by the formula \begin{align*}v=25t-80\end{align*}, where the time is expressed in seconds and the velocity is expressed in feet per second. Find the times for which the magnitude of the velocity is greater than or equal to 60 feet per second.

Solution

The magnitude of the velocity is the absolute value of the velocity. If the velocity is \begin{align*}25t-80\end{align*} feet per second, then its magnitude is \begin{align*}|25t-80|\end{align*} feet per second. We want to find out when that magnitude is greater than or equal to 60, so we need to solve \begin{align*}|25t-80| \ge 60\end{align*} for \begin{align*}t\end{align*}.

First we have to split it up: \begin{align*}25t-80 \ge 60\end{align*} or \begin{align*}25t-80 \le -60\end{align*}

Then solve: \begin{align*}25t \ge 140\end{align*} or \begin{align*}25t \le 20\end{align*}

\begin{align*}t \ge 5.6\end{align*} or \begin{align*}t \le 0.8\end{align*}

The magnitude of the velocity is greater than 60 ft/sec for times less than 0.8 seconds and for times greater than 5.6 seconds.

When \begin{align*}t = 0.8 \ seconds, \ v=25(0.8)-80 = -60 \ ft/sec\end{align*}. The magnitude of the velocity is 60 ft/sec. (The negative sign in the answer means that the object is moving backwards.)

When \begin{align*}t = 5.6 \ seconds, \ v=25(5.6)-80 = 60 \ ft/sec\end{align*}.

To find where the magnitude of the velocity is greater than 60 ft/sec, check some arbitrary values in each of the following time intervals: \begin{align*}t \le 0.8, \ 0.8 \le t \le 5.6\end{align*} and \begin{align*}t \ge 5.6\end{align*}.

Check \begin{align*}t = 0.5: \ v=25(0.5) -80 = -67.5 \ ft/sec\end{align*}

Check \begin{align*}t = 2: \ v = 25(2)-80 =-30 \ ft/sec\end{align*}

Check \begin{align*}t = 6: \ v=25(6)-80 = -70 \ ft/sec\end{align*}

You can see that the magnitude of the velocity is greater than 60 ft/sec only when \begin{align*}t \ge 5.6\end{align*} or when \begin{align*}t \le 0.8\end{align*}.

## Further Resources

For a multimedia presentation on absolute value equations and inequalities, see http://www.teachertube.com/viewVideo.php?video_id=124516.

## Lesson Summary

• The absolute value of a number is its distance from zero on a number line.
• \begin{align*}|x|=x\end{align*} if \begin{align*}x\end{align*} is not negative, and \begin{align*}|x|=-x\end{align*} if \begin{align*}x\end{align*} is negative.
• An equation or inequality with an absolute value in it splits into two equations, one where the expression inside the absolute value sign is positive and one where it is negative. When the expression within the absolute value is positive, then the absolute value signs do nothing and can be omitted. When the expression within the absolute value is negative, then the expression within the absolute value signs must be negated before removing the signs.
• Inequalities of the type \begin{align*}|x| can be rewritten as “\begin{align*}-a < x < a\end{align*}.”
• Inequalities of the type \begin{align*}|x|>b\end{align*} can be rewritten as “\begin{align*}x < -b\end{align*} or \begin{align*}x > b\end{align*}.”

## Review Questions

Evaluate the absolute values.

1. \begin{align*}|250|\end{align*}
2. \begin{align*}|-12|\end{align*}
3. \begin{align*}\left | -\frac{2}{5} \right |\end{align*}
4. \begin{align*}\left | \frac{1}{10} \right |\end{align*}

Find the distance between the points.

1. 12 and -11
2. 5 and 22
3. -9 and -18
4. -2 and 3

Solve the absolute value equations and interpret the results by graphing the solutions on the number line.

1. \begin{align*}|x-5|=10\end{align*}
2. \begin{align*}|x+2|=6\end{align*}
3. \begin{align*}|5x-2|=3\end{align*}
4. \begin{align*}|x-4|=-3\end{align*}

Graph the absolute value functions.

1. \begin{align*}y=|x+3|\end{align*}
2. \begin{align*}y=|x-6|\end{align*}
3. \begin{align*}y=|4x+2|\end{align*}
4. \begin{align*}y= \left | \frac{x}{3}-4 \right |\end{align*}

Solve the following inequalities and show the solution graph.

1. \begin{align*}|x| \le 6\end{align*}
2. \begin{align*}|x| > 3.5\end{align*}
3. \begin{align*}|x| < 12\end{align*}
4. \begin{align*}|x| > 10\end{align*}
5. \begin{align*}|7x| \ge 21\end{align*}
6. \begin{align*}|x-5| > 8\end{align*}
7. \begin{align*}|x+7| < 3\end{align*}
8. \begin{align*}\left | x-\frac{3}{4} \right | \le \frac{1}{2}\end{align*}
9. \begin{align*}|2x-5| \ge 13\end{align*}
10. \begin{align*}|5x+3| < 7\end{align*}
11. \begin{align*}\left | \frac{x}{3}-4 \right | \le 2\end{align*}
12. \begin{align*}\left | \frac{2x}{7}+9 \right | > \frac{5}{7}\end{align*}
1. How many solutions does the inequality \begin{align*}|x| \le 0\end{align*} have?
2. How about the inequality \begin{align*}|x| \ge 0\end{align*}?
13. A company manufactures rulers. Their 12-inch rulers pass quality control if they are within \begin{align*}\frac{1}{32} \ inches\end{align*} of the ideal length. What is the longest and shortest ruler that can leave the factory?
14. A three month old baby boy weighs an average of 13 pounds. He is considered healthy if he is at most 2.5 lbs. more or less than the average weight. Find the weight range that is considered healthy for three month old boys.

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