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# Absolute Value Inequalities

## Inequalities with solution sets 'between' and 'above or below' certain values

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Absolute Value Inequalities

### 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*}.

#### Solving Inequalities

Solve the following inequalities and show the solution graph.

a) \begin{align*}|x|<5\end{align*}

\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*}

\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.

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

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.”

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

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

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

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

#### Solve Real-World Problems Using Absolute Value Inequalities

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

#### Real-World Application: Velocity

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.

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*}.

### Example

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

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

### Review

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*}
13. .
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*}?
14. 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?
15. 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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### Vocabulary Language: English

TermDefinition
Absolute Value Linear inequalities Absolute value linear inequalities can have one of four forms $|ax + b| > c, |ax + b| < c, |ax + b| \ge c$, or $|ax + b| \le c$. Absolute value linear inequalities have two related inequalities. For example for $|ax+b|>c$, the two related inequalities are $ax + b > c$ and $ax + b < -c$.

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