<meta http-equiv="refresh" content="1; url=/nojavascript/"> Absolute Value Inequalities ( Read ) | Algebra | CK-12 Foundation
You are viewing an older version of this Concept. Go to the latest version.

# Absolute Value Inequalities

%
Best Score
Practice Absolute Value Inequalities
Best Score
%

# Solving Absolute Value Inequalities

The tolerance for the weight of a volleyball is 2.6 grams. If the average volleyball weighs 260 grams, what is the range of weights for a volleyball?

### Guidance

Like absolute value equations, absolute value inequalities also will have two answers. However, they will have a range of answers, just like compound inequalities.

$|x|>1$ This inequality will have two answers, when $x$ is 1 and when $-x$ is 1. But, what about the inequality sign? The two possibilities would be:

Notice in the second inequality, we did not write $x>-1$ . This is because what is inside the absolute value sign can be positive or negative. Therefore, if $x$ is negative, then $-x>1$ . It is a very important difference between the two inequalities. Therefore, for the first solution, we leave the inequality sign the same and for the second solution we need to change the sign of the answer AND flip the inequality sign.

#### Example A

Solve $|x+2| \le 10$ .

Solution: There will be two solutions, one with the answer and sign unchanged and the other with the inequality sign flipped and the answer with the opposite sign.

$& \qquad \quad |x+2| \le 10\\& \qquad \qquad \swarrow \searrow\\& x+2 \le 10 \qquad x+2 \ge -10\\& \quad \ \ x \le 8 \qquad \qquad \ x \ge -12$

Test a solution, $x = 0:$

$|0+2| & \le 10\\|2| & \le 10$

When graphing this inequality, we have

Notice that this particular absolute value inequality has a solution that is an “and” inequality because the solution is between two numbers.

If $|ax+b| < c$ where $a> 0$ and $c> 0$ , then $-c .

If $|ax+b| \le c$ where $a> 0$ and $c> 0$ , then $-c \le ax+b \le c$ .

If $|ax+b| > c$ where $a> 0$ and $c> 0$ , then $ax+b<-c$ or $ax+b>c$ .

If $|ax+b| \ge c$ where $a> 0$ and $c> 0$ , then $ax+b \le -c$ or $ax+b \ge c$ .

If $a< 0$ , we will have to divide by a negative and have to flip the inequality sign. This would change the end result. If you are ever confused by the rules above, always test one or two solutions and graph it.

#### Example B

Solve and graph $|4x-3|>9$ .

Solution: Break apart the absolute value inequality to find the two solutions.

$& \qquad \ \ \ |4x-3|>9\\& \qquad \qquad \swarrow \searrow\\& 4x-3>9 \quad \ 4x-3<-9\\& \quad \ \ 4x>12 \qquad \ 4x<-6\\& \qquad x>3 \qquad \quad \ \ x<-\frac{3}{2}$

Test a solution, $x = 5:$

$|4(5)-3|& >9\\|20-3| & >9 \\17 & >9$

The graph is:

#### Example C

Solve $|-2x+5|<11$ .

Solution: In this example, the rules above do not apply because $a< 0$ . At first glance, this should become an “and” inequality. But, because we will have to divide by a negative number, $a$ , the answer will be in the form of an “or” compound inequality. We can still solve it the same way we have solved the other examples.

$& \qquad \quad \ |-2x+5|<11\\& \qquad \qquad \quad \ \ \swarrow \searrow\\& -2x+5<11 \quad -2x+5>-11\\& \quad \ \ -2x<6 \qquad \quad \ -2x>-16\\& \qquad \quad \ x>-3 \qquad \qquad \ x<-8$

The solution is less than -8 or greater than -3.

The graph is:

When $a < 0$ for an absolute value inequality, it switches the results of the rules listed above.

Intro Problem Revisit Set up an absolute value inequality. w is the range of weights of the volleyball.

$& \qquad \qquad |w-260| \le 2.6\\& \qquad \qquad \quad \swarrow \searrow\\& w-260 \le 2.6 \qquad w-260 \ge -2.6\\& \quad \ \ w \le 262.6 \qquad \qquad \ w \ge 257.4$

So, the range of the weight of a volleyball is $257.4 \le w \le 262.6$ grams.

### Guided Practice

1. Is $x = -4$ a solution to $|15-2x|>9$ ?

2. Solve and graph $\bigg |\frac{2}{3}x+5 \bigg | \le 17$ .

1. Plug in -4 for $x$ to see if it works.

$|15-2(-4)| > 9\\|15+8| > 9\\|23| > 9\\23 > 9$

Yes, -4 works, so it is a solution to this absolute value inequality.

2. Split apart the inequality to find the two answers.

$& \qquad \quad \ \bigg | \frac{2}{3}x+5\bigg | \le 17\\& \qquad \qquad \ \ \ \swarrow \searrow\\& \bigg | \frac{2}{3}x+5\bigg | \le 17 \qquad \frac{2}{3}x+5 \ge -17\\& \qquad \ \frac{2}{3}x \le 12 \qquad \quad \ \ \frac{2}{3}x \ge -22\\& \qquad \quad x \le 12 \cdot \frac{3}{2} \qquad \quad x \ge -22 \cdot \frac{3}{2}\\& \qquad \quad x \le 18 \qquad \qquad \ x \ge -33$

Test a solution, $x = 0:$

$\bigg | \frac{2}{3}(0)+5\bigg | & \le 17\\|5| & \le 17 \\5 & \le 17$

### Practice

Determine if the following numbers are solutions to the given absolute value inequalities.

1. $|x-9|>4;10$
2. $\bigg | \frac{1}{2} x-5 \bigg | \le 1;8$
3. $|5x+14| \ge 29;-8$

Solve and graph the following absolute value inequalities.

1. $|x+6|>12$
2. $|9-x| \le 16$
3. $|2x-7| \ge 3$
4. $|8x-5|<27$
5. $\bigg | \frac{5}{6}x+1 \bigg |>6$
6. $|18-4x| \le 2$
7. $\bigg | \frac{3}{4}x-8 \bigg |>13$
8. $|6-7x| \le 34$
9. $|19+3x| \ge 46$

Solve the following absolute value inequalities. $a$ is greater than zero.

1. $|x-a|>a$
2. $|x+a| \le a$
3. $|a-x| \le a$