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# 6.6: Absolute Value Inequalities

Difficulty Level: At Grade Created by: CK-12

An absolute value inequality is a combination of two concepts: absolute values and linear inequalities. Therefore, to solve an absolute value inequality, you must use the problem-solving methods of each concept.

• To solve an absolute value equation, use the definition.
• If d=|xa|\begin{align*}d=|x-a|\end{align*}, then xa=d\begin{align*}x-a=d\end{align*} OR xa=d\begin{align*}x-a= -d\end{align*}.
• To solve a linear inequality, use the concepts learned in Lessons 2 and 3 of this chapter.
• Remember, when dividing by a negative, the inequality symbol must be reversed!

Let’s begin by looking at an example.

|x|3\begin{align*}|x| \le 3\end{align*}

Since |x|\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 3x3\begin{align*}-3 \le x\le 3\end{align*}.

Below is a second example.

|x|>2\begin{align*}|x|>2\end{align*}

Since the absolute value of x\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.

In general, the solutions to absolute value inequalities take two forms:

1. If |x|<a\begin{align*}|x|, then x<a\begin{align*}x or x>a\begin{align*}x>-a\end{align*}.
2. If |x|>a\begin{align*}|x|>a\end{align*}, then x>a\begin{align*}x>a\end{align*} or x<a\begin{align*}x<-a\end{align*}.

Example 1: Solve |x+5|>7\begin{align*}|x+5|>7\end{align*}.

Solution: This equation fits situation 2. Therefore, x+5>7\begin{align*}x+5>7\end{align*} OR x+5<7\begin{align*}x+5<-7\end{align*}

Solve each inequality separately.

x+5x>7>2x+5<7  x<12\begin{align*}x+5&>7 && x+5<-7 \\ x&>2 && \quad \ \ x<-12\end{align*}

The solutions are all values greater than two or less than –12.

Example: The velocity of an object is given by the formula v=25t80\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 when the velocity is greater than or equal to 60 feet per second.

Solution: We want to find the times when the velocity is greater than or equal to 60 feet per second. Using the formula for velocity v=25t80\begin{align*}v=25t-80\end{align*} and substituting the appropriate values, we obtain the absolute value inequality |25t80|60\begin{align*}|25t-80|\ge 60\end{align*}

This is an example like case 2. Separate and solve.

25t8060\begin{align*}25t-80 \ge 60\end{align*} or 25t8060\begin{align*}25t-80 \le -60\end{align*}

25t140\begin{align*}25t \ge 140\end{align*} or 25t20\begin{align*}25t \le 20\end{align*}

t5.6\begin{align*}t \ge 5.6\end{align*} or t0.8\begin{align*}t \le 0.8\end{align*}

Multimedia Links: For more assistance with graphing absolute value inequalities, visit this YouTube video:

Or you can also visit this link on TheMathPage: http://www.themathpage.com/alg/absolute-value.htm.

## Practice Set

Sample explanations for some of the practice exercises below are available by viewing the following video. Note that there is not always a match between the number of the practice exercise in the video and the number of the practice exercise listed in the following exercise set.  However, the practice exercise is the same in both.

1. You are asked to solve |a+1|4\begin{align*}|a+1|\le 4\end{align*}. What two inequalities does this separate into?

In 2 – 21, solve the inequality and show the solution graph.

1. |x|6\begin{align*}|x| \le 6\end{align*}
2. 4|a+4|\begin{align*}4 \le |a+4|\end{align*}
3. |x|>3.5\begin{align*}|x| > 3.5\end{align*}
4. 6>|10b+6|\begin{align*}6 > |10b+6|\end{align*}
5. |x|<12\begin{align*}|x| < 12\end{align*}
6. w10<2\begin{align*}\left |\frac{w}{10}\right | < 2\end{align*}
7. x56\begin{align*}\left |\frac{x}{5}\right | \le 6\end{align*}
8. |7x|21\begin{align*}|7x| \ge 21\end{align*}
9. |6c+5|<47\begin{align*}|6c+5| < 47\end{align*}
10. |x5|>8\begin{align*}|x-5| > 8\end{align*}
11. |x+7|<3\begin{align*}|x+7| < 3\end{align*}
12. |x34|12\begin{align*}|x-\frac{3}{4}| \le \frac{1}{2}\end{align*}
13. |2x5|13\begin{align*}|2x-5| \ge 13\end{align*}
14. |5x+3|<7\begin{align*}|5x+3|<7\end{align*}
15. x342\begin{align*}\left |\frac{x}{3}-4\right | \le 2\end{align*}
16. 2x7+9>57\begin{align*}\left |\frac{2x}{7}+9\right | > \frac{5}{7}\end{align*}
17. |6t+3|+918\begin{align*}|-6t+3|+9 \ge 18\end{align*}
18. |9p+5|>23\begin{align*}|9p+5|>23\end{align*}
19. |2s4|6\begin{align*}|-2s-4|\le 6\end{align*}
20. |10m5|8>5\begin{align*}\frac{|10m-5|}{8}>5\end{align*}
21. A three-month-old baby boy weighs an average of 13 pounds. He is considered healthy if he is 2.5 pounds more or less than the average weight. Find the weight range that is considered healthy for three-month-old boys.

Mixed Review

1. Solve |7u|=77\begin{align*}|7u|=77\end{align*}.
2. A map has a scale of 2 inch=125 miles\begin{align*}2 \ inch=125 \ miles\end{align*}. How far apart would two cities be on the map if the actual distance is 945 miles?
3. Determine the domain and range: {(9,0),(6,0),(4,0),(0,0),(3,0),(5,0)}\begin{align*}\left \{(-9,0),(-6,0),(-4,0),(0,0),(3,0),(5,0)\right \}\end{align*}.
4. Is the relation in question #25 a function? Explain your reasoning.
5. Consider the problem 3(2x7)=100\begin{align*}3(2x-7)=100\end{align*}. Lei says the first step to solving this equation is to use the Distributive Property to cancel the parentheses. Hough says the first step to solving this equation is to divide by 3. Who is right? Explain your answer.
6. Graph 4x+y=6\begin{align*}4x+y=6\end{align*} using its intercepts.
7. Write 330\begin{align*}\frac{3}{30}\end{align*} as a percent. Round to the nearest hundredth.
8. Simplify 523÷718\begin{align*}-5\frac{2}{3} \div \frac{71}{8}\end{align*}.

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