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# 2.14: Algebraic Solutions to Absolute Value Inequalities

Difficulty Level: At Grade Created by: CK-12

A ball is fired from the cannon during the 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 s=|32t+150|\begin{align*}s =|-32t+150|\end{align*}, where \begin{align*}s\end{align*} is the speed measure in ft/sec and t is the time in seconds. Calculate the times when the speed is greater than 86 ft/sec.

### Guidance

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, if you look at 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 inequality expressions 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*}
Equation 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*}
Equation 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. It is also important to remember, as you found with absolute value linear equations, that if \begin{align*}c < 0\end{align*}, there is no solution.

#### Example A

Solve for the absolute value inequality \begin{align*}|g+5|<3\end{align*}.

Solution: \begin{align*}|g+5|<3\end{align*}

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

Solutions \begin{align*}g <-2\end{align*} and \begin{align*}g>-8\end{align*}.

#### Example B

Solve for the absolute value inequality \begin{align*}\big|j-\frac{1}{2}\big|>2\end{align*}.

Solution:

\begin{align*}\big|j-\frac{1}{2}\big|>2\end{align*}

\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 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 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*}

Solutions \begin{align*}j > \frac{3}{2}\end{align*} and \begin{align*}j < \frac{-1}{2}\end{align*}.

#### Example C

Solve for the absolute value inequality \begin{align*}|t+1|-3 \ge 2\end{align*}.

Solution:

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

\begin{align*}t+1-3 & \ge 2\\ t-2 & \ge 2 && \text{Combine constant terms on left side of inequality sign}\\ t-2{\color{red}+2} & \ge 2 {\color{red}+2} && \text{Add 2 to sides to isolate and solve for the variable}\\ t \ge 4\\ & OR\\ t+1-3 & \le -2\\ t-2 & \le -2 && \text{Combine constant terms on left side of inequality sign}\\ t-2{\color{red}+2} & \le -2 {\color{red}+2} && \text{Add 2 to sides to isolate and solve for the variable}\\ t & \le 0\end{align*}

Solutions \begin{align*}t \ge 4\end{align*} and \begin{align*}t<0\end{align*}.

#### Concept Problem Revisited

A ball is fired from the cannon during the 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 t is the time in seconds. Calculate the times when the speed is greater 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 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 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 &< 7.4\end{align*}

Therefore when \begin{align*}t >2\end{align*} or \begin{align*}t <7.4\end{align*}, the speed is greater than 64 ft/sec.

### Vocabulary

Absolute Value Linear Inequality
Absolute Value Linear inequalities can have one of four forms: \begin{align*}|ax + b| > c, |ax + b| < c, |ax + b| \ge c\end{align*}, or \begin{align*}|ax + b| \le c\end{align*}. Absolute value linear inequalities have two related inequalities. For example for \begin{align*}|ax+b|>c\end{align*}, the two related inequalities are \begin{align*}ax + b > c\end{align*} and \begin{align*}ax + b > -c\end{align*}.
Linear Inequality
Linear inequalities can have one of four forms: \begin{align*}ax + b > c, ax + b < c, ax + b \ge c\end{align*}, or \begin{align*}ax + b \le c\end{align*}. In other words, the left side no longer equals the right side, it is less than, greater than, less than or equal to, or greater than or equal to.

### Guided Practice

1. Solve for the solution set to the inequality \begin{align*}|x-1| \ge 9\end{align*}.

2. Solve for the solution set to the inequality \begin{align*}|-2w+7|<23\end{align*}.

3. Solve for the solution set to the inequality \begin{align*}|-4+2b|+3 \le 21\end{align*}.

1. \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*}

Therefore solution set is \begin{align*}x \ge 10,x \le -8\end{align*}.

2. \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*}

Therefore solution set is \begin{align*}w >-8,w<15\end{align*}.

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

\begin{align*}-4+2b+3 &\le 21\\ 2b-1 & \le 21 && (\text{Combine constant terms on left side of inequality sign})\\ 2b-1{\color{red}+1} & \le 21{\color{red}+1} && (\text{Add 1 to both sides to isolate the variable})\\ 2b & \le 22 && (\text{Simplify})\\ \frac{2b}{{\color{red}2}} &\le \frac{22}{{\color{red}2}} && (\text{Divide both sides by 2 to solve for the variable})\\ b & \le 11\\ & OR\\ -4+2b+3 & \ge -21\\ 2b-1 & \ge -21 && (\text{Combine constant terms on left side of inequality sign})\\ 2b-1{\color{red}+1} & \ge -21{\color{red}+1} && (\text{Add 1 to both sides to isolate the variable})\\ 2b & \ge -20 && (\text{Simplify})\\ \frac{2b}{{\color{red}2}} &\ge \frac{-20}{{\color{red}2}} && (\text{Divide both sides by 2 to solve for the variable})\\ b & \ge -10\end{align*}

Therefore solution set is \begin{align*}b \le 11,b \ge -10\end{align*}.

### Practice

Find the solution sets for the variable in 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*}

Find the solution sets for the variable in each of the following absolute value linear inequalities.

1. \begin{align*}|s+2|-5>8\end{align*}
2. \begin{align*}|10+8w|-2<16\end{align*}
3. \begin{align*}|2q+1|-5 \le 7\end{align*}
4. \begin{align*}\big |\frac{1}{3}(g-2) \big |<4\end{align*}
5. \begin{align*}|-2(e+4)|>17\end{align*}

Find the solution sets for the variable in each of the following absolute value linear inequalities.

1. \begin{align*}|-5x-3(2x-1)|>3\end{align*}
2. \begin{align*}|2(a-1.2)|\ge 5.6\end{align*}
3. \begin{align*}|-2(r+3.1)| \le -1.4\end{align*}
4. \begin{align*}\big|\frac{3}{4}(m-3)\big| \le 8\end{align*}
5. \begin{align*}\big|-2\left(e-\frac{3}{4}\right)\big| \ge 3\end{align*}

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Dec 19, 2012