# Absolute Value Inequalities

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

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Algebraic Solutions to Absolute Value Inequalities

A ball is fired from a cannon during Independence Day celebrations. It is fired directly into the air with an initial velocity of 150 ft/sec. The speed of the cannon ball can be calculated using the formula \begin{align*}s =|-32t+150|,\end{align*} where \begin{align*}s\end{align*} is the speed measure in ft/sec and \begin{align*}t\end{align*} is the time in seconds. Calculate the times when the speed is less than 86 ft/sec.

### Algebraic Solutions to Absolute Value Inequalities

You have learned that a linear inequality is of the form \begin{align*}ax + b > c, \ ax + b < c, \ ax + b \ge c\end{align*}, or \begin{align*}ax + b \le c.\end{align*} Linear inequalities, unlike linear equations, have more than one solution. They have a solution set. For example, consider the linear inequality \begin{align*}x + 3 > 5\end{align*}. You know that \begin{align*}2 + 3\end{align*} is equal to 5, therefore the solution set could be any number greater than 2.

Recall that when solving absolute value linear equations, you have to solve for the two related equations. Remember that for \begin{align*}|ax+b|=c\end{align*}, you had to solve for \begin{align*}ax+b=c\end{align*} and \begin{align*}ax+b=-c\end{align*}. The same is true for linear inequalities. If you have an absolute value linear inequality, you would need to solve for the two related linear inequalities.

The table below shows the four types of absolute value linear inequalities and the two related inequalities required to be solved for each one.

Absolute Value Inequality \begin{align*}|ax+b|>c\end{align*} \begin{align*}|ax+b| \begin{align*}|ax+b|\ge c\end{align*} \begin{align*}|ax+b|\le c\end{align*}
Inequality 1 \begin{align*}ax+b>c\end{align*} \begin{align*}ax+b \begin{align*}ax+b \ge c\end{align*} \begin{align*}ax+b \le c\end{align*}
Inequality 2 \begin{align*}ax+b<-c\end{align*} \begin{align*}ax+b>-c\end{align*} \begin{align*}ax+b \le -c\end{align*} \begin{align*}ax+b \ge -c\end{align*}

Remember the rules to algebraically solve for the variable remain the same as you have used before.

#### Let's solve the following absolute value inequalitites:

1.  \begin{align*}|g+5|<3\end{align*}

Set up and solve two inequalities:

\begin{align*}g+5 &< 3\\ g+5{\color{red}-5} &< 3{\color{red}-5} && \text{Subtract 5 from both sides to isolate the variable}\\ g &< -2\\ & OR\\ g+5 &>-3\\ g+5{\color{red}-5} &> -3{\color{red}-5} && \text{Subtract 5 from both sides to isolate the variable}\\ g &> -8\end{align*}

Solution: \begin{align*}-8

1.  \begin{align*}\left| \ j-\frac{1}{2} \, \right| >2\end{align*}

Set up and solve two inequalities:

\begin{align*}j-\frac{1}{2} &>2\\ \left({\color{red}\frac{2}{2}}\right)j-\frac{1}{2} & > \left({\color{red}\frac{2}{2}}\right)2 && \text{Multiply to get a common denominator (LCD} = 2)\\ \frac{2j}{2}-\frac{1}{2} & > \frac{2}{2} && \text{Simplify}\\ 2j-1 &> 2 && \text{Simplify}\\ 2j-1{\color{red}+1} &>2 {\color{red}+1} && \text{Add 1 to both sides to isolate the variable}\\ 2j &> 3 && \text{Simplify}\\ \frac{2j}{{\color{red}2}}&>\frac{3}{{\color{red}2}} && \text{Divide by 2 to solve for the variable.}\\ j &>\frac{3}{2}\\ & OR\\ j-\frac{1}{2} &<-2\\ \left({\color{red}\frac{2}{2}}\right)j-\frac{1}{2} &<\left({\color{red}\frac{2}{2}}\right)(-2) && \text{Multiply to get a common denominator (LCD} = 2)\\ \frac{2j}{2}-\frac{1}{2} &< \frac{-2}{2} && \text{Simplify}\\ 2j-1 &< -2 && \text{Simplify}\\ 2j-1{\color{red}+1} &<-2{\color{red}+1} && \text{Add 1 to both sides to isolate the variable}\\ 2j &< -1 && \text{Simplify}\\ \frac{2j}{{\color{red}2}}&<\frac{-1}{{\color{red}2}} && \text{Divide by 2 to solve for the variable.}\\ j &<\frac{-1}{2}\end{align*}

Solution: \begin{align*}j > \frac{3}{2}\end{align*} or \begin{align*}j < \frac{-1}{2}\end{align*}

1.  \begin{align*}|t+1|-3 \ge 2\end{align*}

First, isolate the absolute value part of the inequality:

\begin{align*}|t+1|-3 & \ge 2\\ |t+1|-3{\color{red}+3} & \ge 2 {\color{red}+3} \\ |t+1| & \ge 5 \end{align*}

Now, set up and solve the two inequalities:

\begin{align*}t+1 & \ge 5\\ t+1{\color{red}-1} & \ge 5 {\color{red}-1} \\ t \ge 4\\ & OR\\ t+1 & \le -5\\ t+1{\color{red}-1} & \le -5 {\color{red}-1} \\ t & \le -6\end{align*}

Solution: \begin{align*}t \ge 4\end{align*} or \begin{align*}t\le -6\end{align*}

### Examples

#### Example 1

Earlier, you were told that a ball is fired from a cannon during Independence Day celebrations. It is fired directly into the air with an initial velocity of 150 ft/sec. The speed of the cannon ball can be calculated using the formula \begin{align*}s =|-32t+150|\end{align*}, where \begin{align*}s\end{align*} is the speed measure in ft/sec and \begin{align*}t\end{align*} is the time in seconds. Calculate the times when the speed is less than 86 ft/sec.

\begin{align*}86 > |-32t+150|\end{align*}

\begin{align*}86 & > -32t+150\\ 86 {\color{red}-150} &> -32t+150{\color{red}-150} && \text{Subtract 150 from both sides to isolate the variable}\\ -64 & > -32t && \text{Simplify}\\ \frac{-64}{{\color{red}-32}} & < \frac{-32t}{{\color{red}-32}} && \text{Divide by -32 to solve for the variable. Remember when}\\ & && \quad \ \text{dividing by a negative number to reverse the sign of the inequality.}\\ t &> 2\\ & OR\\ -86 & < -32t+150\\ -86 {\color{red}-150} & < -32t+150{\color{red}-150} && \text{Subtract 150 from both sides to isolate the variable}\\ -236 & < -32t && \text{Simplify}\\ \frac{-236}{{\color{red}-32}} & > \frac{-32t}{{\color{red}-32}} && \text{Divide by -32 to solve for the variable. Remember when}\\ & && \quad \ \text{dividing by a negative number to reverse the sign of the inequality.}\\ t &< 7.375\end{align*}

Therefore when \begin{align*}2 the speed is greater than 86 ft/sec.

#### Example 2

Solve the inequality:

\begin{align*}|x-1| \ge 9\end{align*}

\begin{align*}|x-1| \ge 9\end{align*}

\begin{align*}x-1 & \ge 9\\ x-1{\color{red}+1} & \ge 9{\color{red}+1} && (\text{Add 1 to both sides to isolate and solve for the variable})\\ x & \ge 10\\ & OR\\ x-1 & \le -9\\ x-1{\color{red}+1} & \le -9{\color{red}+1} && (\text{Add 1 to both sides to isolate and solve for the variable})\\ x & \le -8\end{align*}

Solution: \begin{align*}x \ge 10\end{align*} or \begin{align*}x \le -8\end{align*}.

#### Example 3

Solve the inequality:

\begin{align*}|-2w+7|<23\end{align*}

\begin{align*}|-2w+7| < 23\end{align*}

\begin{align*}-2w+7 &< 23\\ -2w+7{\color{red}-7} &< 23{\color{red}-7} && (\text{Subtract 7 from both sides to get variables on same side})\\ -2w &< 16 && (\text{Simplify})\\ \frac{-2w}{{\color{red}-2}} &> \frac{16}{{\color{red}-2}} &&(\text{Divide by -2 to solve for the variable, reverse sign of inequality})\\ w &> -8\\ & OR \\ -2w+7 &> -23\\ -2w+7{\color{red}-7} &> -23{\color{red}-7} && ( \text{Subtract 7 from both sides to get variables on same side})\\ -2w &> -30 && (\text{Simplify})\\ \frac{-2w}{{\color{red}-2}} &< \frac{-30}{{\color{red}-2}} &&(\text{Divide by -2 to solve for the variable, reverse sign of inequality})\\ w &<15\end{align*}

Solution: \begin{align*}-8

#### Example 4

Solve the inequality:

\begin{align*}|-4+2b|+3 \le 21\end{align*}

First, isolate the absolute value part of the inequality:

\begin{align*}|-4+2b|+3 & \le 21\\ |-4+2b|+3{\color{red}-3} & \le 21 {\color{red}-3} \\ |-4+2b| & \le 18 \end{align*}

Now, set up and solve the two inequalities:

\begin{align*}-4+2b & \le 18\\ -4+2b{\color{red}+4} & \le 18 {\color{red}+4} \\ 2b & \le 22\\ b & \le 11\\ & OR\\ -4+2b & \ge -18\\ -4+2b{\color{red}+4} & \ge -18 {\color{red}+4} \\ 2b & \ge -14\\ b & \ge -7\end{align*}

Solution: \begin{align*}-7\le b\le 11\end{align*}

### Review

Solve each of the following absolute value linear inequalities:

1. \begin{align*}|p-16|>10\end{align*}
2. \begin{align*}|r+2|<5\end{align*}
3. \begin{align*}|3-2k|\ge 1\end{align*}
4. \begin{align*}|8-y|>5\end{align*}
5. \begin{align*}8 \ge |5d-2|\end{align*}
6. \begin{align*}|s+2|-5>8\end{align*}
7. \begin{align*}|10+8w|-2<16\end{align*}
8. \begin{align*}|2q+1|-5 \le 7\end{align*}
9. \begin{align*}\big |\frac{1}{3}(g-2) \big |<4\end{align*}
10. \begin{align*}|-2(e+4)|>17\end{align*}
11. \begin{align*}|-5x-3(2x-1)|>3\end{align*}
12. \begin{align*}|2(a-1.2)|\ge 5.6\end{align*}
13. \begin{align*}|-2(r+3.1)| \le 1.4\end{align*}
14. \begin{align*}\big|\frac{3}{4}(m-3)\big| \le 8\end{align*}
15. \begin{align*}\big|-2\left(e-\frac{3}{4}\right)\big| \ge 3\end{align*}

To see the Review answers, open this PDF file and look for section 2.14.

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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 $ax + b < -c$.
Linear Inequality Linear inequalities are inequalities that can be written in one of the following four forms: $ax + b > c, ax + b < c, ax + b \ge c$, or $ax + b \le c$.