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# Absolute Value Inequalities

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

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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?

### Absolute Value Inequalities

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

Consider the inequality |x|>1.\begin{align*}|x|>1.\end{align*} If this were an equation, it would have two answers: one when x\begin{align*}x\end{align*} is 1 and the other when -x\begin{align*}\text{-}x\end{align*} is 1, but how does the inequality sign affect it? As an inequality, the two possibilities would be:

Notice in the second inequality, we did not write x>1\begin{align*}x>-1\end{align*}. This is because what is inside the absolute value sign can be positive or negative. Therefore, if x\begin{align*}x\end{align*} is negative, then x>1\begin{align*}-x>1\end{align*}. 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.

Let's solve the following absolute value inequalities.

1. Solve |x+2|10.\begin{align*}|x+2| \le 10.\end{align*}

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.

|x+2|10x+210x+210  x8 x12\begin{align*}& \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\end{align*}

Test a solution, x=0:\begin{align*}x = 0:\end{align*}

|0+2||2|1010\begin{align*}|0+2| & \le 10\\ |2| & \le 10 \end{align*}

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\begin{align*}|ax+b| < c\end{align*}, then c<ax+b<c\begin{align*}-c < ax+b < c\end{align*}.

If |ax+b|c\begin{align*}|ax+b| \le c\end{align*}, then cax+bc\begin{align*}-c \le ax+b \le c\end{align*}.

If |ax+b|>c\begin{align*}|ax+b| > c\end{align*}, then ax+b<c\begin{align*}ax+b<-c\end{align*} or ax+b>c\begin{align*}ax+b>c\end{align*}.

If |ax+b|c\begin{align*}|ax+b| \ge c\end{align*}, then ax+bc\begin{align*}ax+b \le -c\end{align*} or ax+bc\begin{align*}ax+b \ge c\end{align*}.

If you are ever confused by the rules above, you can always test one or two solutions and graph it.

1. Solve and graph |4x3|>9.\begin{align*}|4x-3|>9.\end{align*}

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

|4x3|>94x3>9 4x3<9  4x>12 4x<6x>3  x<32\begin{align*} & \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}\end{align*}

Test a solution, x=5:\begin{align*}x = 5:\end{align*}

|4(5)3||203|17>9>9>9\begin{align*}|4(5)-3|& >9\\ |20-3| & >9 \\ 17 & >9\end{align*}

The graph is:

1. Solve |2x+5|<11.\begin{align*}|-2x+5|<11.\end{align*}

Given the rules above, this will become an “and” inequality, so the solution will be a range between two values.

|2x+5|<11  2x+5<112x+5>11  2x<6 2x>16 x>3 x<8\begin{align*}& \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\end{align*}

The solution is x\begin{align*}x\end{align*} is greater than 3\begin{align*}-3\end{align*} and less than 8\begin{align*}8\end{align*}. In other words, the solution is 3<x<8\begin{align*}-3 < x < 8\end{align*}.

The graph is:

### Examples

#### Example 1

Earlier, you were asked to find the range of weights for a volleyball.

Set up an absolute value inequality where w\begin{align*}w\end{align*} is the range of weights of the volleyball.

|w260|2.6w2602.6w2602.6  w262.6 w257.4\begin{align*}& \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\end{align*}

So, the range of the weight of a volleyball is 257.4w262.6\begin{align*} 257.4 \le w \le 262.6 \end{align*} grams.

#### Example 2

Is x=-4\begin{align*}x = \text{-}4\end{align*} a solution to |152x|>9?\begin{align*}|15-2x|>9?\end{align*}

Plug in -4 for x\begin{align*}x\end{align*} to see if it works.

|152(4)|>9|15+8|>9|23|>923>9\begin{align*}|15-2(-4)| > 9\\ |15+8| > 9\\ |23| > 9\\ 23 > 9\end{align*}

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

#### Example 3

Solve and graph 23x+517.\begin{align*}\bigg |\frac{2}{3}x+5 \bigg | \le 17.\end{align*}

Split apart the inequality to find the two answers.

23x+517   23x+51723x+517 23x12  23x22x1232x2232x18 x33\begin{align*}& \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\end{align*}

Test a solution, x=0:\begin{align*}x = 0:\end{align*}

23(0)+5|5|5171717\begin{align*}\bigg | \frac{2}{3}(0)+5\bigg | & \le 17\\ |5| & \le 17 \\ 5 & \le 17\end{align*}

### Review

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

1. |x9|>4;10\begin{align*}|x-9|>4;10\end{align*}
2. 12x51;8\begin{align*}\bigg | \frac{1}{2} x-5 \bigg | \le 1;8\end{align*}
3. |5x+14|29;8\begin{align*}|5x+14| \ge 29;-8\end{align*}

Solve and graph the following absolute value inequalities.

1. |x+6|>12\begin{align*}|x+6|>12\end{align*}
2. |9x|16\begin{align*}|9-x| \le 16\end{align*}
3. |2x7|3\begin{align*}|2x-7| \ge 3\end{align*}
4. |8x5|<27\begin{align*}|8x-5|<27\end{align*}
5. 56x+1>6\begin{align*}\bigg | \frac{5}{6}x+1 \bigg |>6\end{align*}
6. |184x|2\begin{align*}|18-4x| \le 2\end{align*}
7. \begin{align*}\bigg | \frac{3}{4}x-8 \bigg |>13\end{align*}
8. \begin{align*}|6-7x| \le 34\end{align*}
9. \begin{align*}|19+3x| \ge 46\end{align*}

Solve the following absolute value inequalities. \begin{align*}a\end{align*} is greater than zero.

1. \begin{align*}|x-a|>a\end{align*}
2. \begin{align*}|x+a| \le a\end{align*}
3. \begin{align*}|a-x| \le a\end{align*}

To see the Review answers, open this PDF file and look for section 1.14.

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### Vocabulary Language: English

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 $ax + b < -c$.