2.15: Graphical Solutions to Absolute Value Inequalities
Solve the following inequality and graph the solution on a number line.
\begin{align*}|x+2|\le 3\end{align*}
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Khan Academy Absolute Value Inequalities on a Number Line
Guidance
Recall that you can graph linear inequalities on number lines. For \begin{align*}x > 5\end{align*}
Notice that there is only one solution set and therefore one section of the number line has the region shown in red.
What do you think would happen with absolute value linear inequalities? With absolute value linear inequalities, there are two solution sets. Therefore there can be two sections of the number line showing solutions.
For \begin{align*}|t|>5\end{align*}
The solution is \begin{align*} t>5\end{align*}
For \begin{align*}|t|<5\end{align*}
The solution is \begin{align*} -5<t<5\end{align*}
Graphing the solution set to an absolute value linear inequality gives you the same visual representation as you had when graphing the solution set to linear inequalities. The same rules apply when graphing absolute values of linear inequalities on a real number line. Once the solution is found, the open circle is used for absolute value inequalities containing the symbols > and <. The closed circle is used for absolute value inequalities containing the symbols \begin{align*}\le\end{align*}
Example A
Represent the solution set to the following inequality on a number line.
\begin{align*}|2x| &\ge 6\\
2x & \ge 6\\
\frac{2x}{{\color{red}2}} & \ge \frac{6}{{\color{red}2}} && (\text{Divide by 2 to isolate and solve for the variable})\\
x & \ge 3 && (\text{Simplify})\\
& OR\\
2x & \le -6\\
\frac{2x}{{\color{red}2}} & \le \frac{-6}{{\color{red}2}} && (\text{Divide by 2 to isolate and solve for the variable})\\
x & \le -3 && (\text{Simplify})\end{align*}
The solution sets are \begin{align*}x \ge 3\end{align*}
Example B
Solve the following inequality and graph the solution on a number line.
\begin{align*}|x+1| &> 3 && (\text{Divide both sides by 2 to solve for the variable})\\
x+1 &>3\\
x+1{\color{red}-1} &> 3{\color{red}-1} && (\text{Subtract 1 from both sides of the inequality sign})\\
x & > 2\\
& OR\\
x+1 &< -3\\
x+1{\color{red}-1} &< -3{\color{red}-1} && (\text{Subtract 1 from both sides of the inequality sign})\\
x & < -4\end{align*}
The solution sets are \begin{align*}x>2\end{align*}
Example C
Solve the following inequality and graph the solution on a number line.
\begin{align*}\bigg |x-\frac{5}{2} \bigg | &< 1\\
x-\frac{5}{2} &< 1\\
\left({\color{red}\frac{2}{2}}\right)x-\frac{5}{2} &< \left(\frac{{\color{red}2}}{{\color{red}2}}\right)1\\
\frac{2x}{2}-\frac{5}{2}&<\frac{2}{2}\\
2x-5&<2 && (\text{Simplify})\\
2x-5{\color{red}+5} &< 2{\color{red}+5} && ( \text{Subtract 5 to isolate the variable})\\
2x &< 7&& (\text{Simplify})\\
\frac{2x}{{\color{red}2}} &< \frac{7}{{\color{red}2}}\\
x &< \frac{7}{2}\\
& OR\\
x-\frac{5}{2} &> -1\\
\left({\color{red}\frac{2}{2}}\right)x-\frac{5}{2} &> \left(\frac{{\color{red}2}}{{\color{red}2}}\right)(-1) && (\text{Multiply to get common denominator (LCD} = 2))\\
\frac{2x}{2}-\frac{5}{2}&<\frac{-2}{2}&& (\text{Simplify})\\
2x-5&>-2 && (\text{Simplify})\\
2x-5{\color{red}+5} &> -2{\color{red}+5} && (\text{Subtract 5 to isolate the variable})\\
2x &> 3&& (\text{Simplify})\\
\frac{2x}{{\color{red}2}} &> \frac{3}{{\color{red}2}} && (\text{Divide both sides by 2 to solve for the variable})\\
x &> \frac{3}{2}\end{align*}
The solution is \begin{align*}\frac{3}{2}<x<\frac{7}{2}\end{align*}
Concept Problem Revisited
Solve the following inequality and graph the solution on a number line.
\begin{align*}|x+2|\le 3\end{align*}
First solve the inequality:
\begin{align*}x+2 &\le 3\\
x+2{\color{red}-2} &\le 3{\color{red}-2} && \text{Subtract 2 from both sides to isolate the variable}\\
x &\le 1 && \text{Simplify}\\
& OR\\
x+2 &\ge -3\\
x+2{\color{red}-2} &\ge -3{\color{red}-2} && \text{Subtract 2 from both sides to isolate the variable}\\
x &\ge -5 && \text{Simplify}\end{align*}
The solution is \begin{align*}-5 \le x \le 1\end{align*}.
Representing on a number line
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*}.
- Number Line
- A number line is a line that matches a set of points and a set of numbers one to one.
Guided Practice
1. Represent the solution set to the inequality \begin{align*}|2x+3|>5\end{align*} on a number line.
2. Represent the solution set to the inequality \begin{align*}|32x-16| \ge 32\end{align*} on a number line.
3. Represent the solution set to the inequality \begin{align*}|x-21.5|>12.5\end{align*} on a number line.
Answers:
1. \begin{align*}|2x+3| >5\end{align*}
\begin{align*}2x+3&>5\\ 2x+3{\color{red}-3}&>5{\color{red}-3} && (\text{Subtract 3 from both sides of the inequality sign})\\ 2x &> 2 && (\text{Simplify})\\ \frac{2x}{{\color{red}2}}&>\frac{2}{{\color{red}2}} && (\text{Divide by 2 to solve for the variable})\\ x &>1\\ & OR\\ 2x+3&<-5\\ 2x+3{\color{red}-3}&<-5{\color{red}-3} && (\text{Subtract 3 from both sides of the inequality sign})\\ 2x &< -8 && (\text{Simplify})\\ \frac{2x}{{\color{red}2}}&>\frac{-8}{{\color{red}2}} && ( \text{Divide by 2 to solve for the variable})\\ x &<-4\end{align*}
The solution sets are \begin{align*}x>1\end{align*} and \begin{align*}x<-4\end{align*}.
2. \begin{align*}|32x-16| \ge 32\end{align*}
\begin{align*}32x-16 &\ge 32\\ 32x-16{\color{red}+16}&\ge 32{\color{red}+16} && (\text{Add 16 to both sides of the inequality sign})\\ 32x &\ge 48&& (\text{Simplify})\\ \frac{32x}{{\color{red}32}} &\ge \frac{48}{{\color{red}32}}&& (\text{Divide by 32 to solve for the variable})\\ x &\ge \frac{3}{2}\\ & OR \\ 32x-16 &\le -32\\ 32x-16{\color{red}+16}&\le -32{\color{red}+16} && (\text{Add 16 to both sides of the inequality sign})\\ 32x &\le -16&& (\text{Simplify})\\ \frac{32x}{{\color{red}32}} &\le \frac{-16}{{\color{red}32}}&& (\text{Divide by 32 to solve for the variable})\\ x &\le -\frac{1}{2}\end{align*}
The solution sets are \begin{align*}x \ge \frac{3}{2}, x \le -\frac{1}{2}\end{align*}.
3. \begin{align*}|x-21.5|>12.5\end{align*}
\begin{align*}x-21.5 &> 12.5\\ x-21.5{\color{red}+21.5}&>12.5{\color{red}+21.5} && (\text{Add 21.5 to both sides to isolate the variable})\\ x &>34 &&(\text{Simplify})\\ & OR\\ x-21.5 &< -12.5\\ x-21.5{\color{red}+21.5}&<-12.5{\color{red}+21.5} && (\text{Add 21.5 to both sides to isolate the variable})\\ x &<9 &&(\text{Simplify})\end{align*}
The solution sets are \begin{align*}x<9,x>34\end{align*}.
Practice
- Represent the solution sets to the absolute value inequality \begin{align*}|3-2x|<3\end{align*} on a number line.
- Represent the solution sets to the absolute value inequality \begin{align*}2\big|\frac{2x}{3}+1\big|\ge 4\end{align*} on a number line.
- Represent the solution sets to the absolute value inequality \begin{align*}\big|\frac{2g-9}{4}\big|<1\end{align*} on a number line.
- Represent the solution sets to the absolute value inequality \begin{align*}\big|\frac{4}{3}x-5\big|\ge 7\end{align*} on a number line.
- Represent the solution sets to the absolute value inequality \begin{align*}|2x+5|+4 \ge 1\end{align*} on a number line.
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Here you'll learn how to represent the solutions of an absolute value inequality on a number line.
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