<meta http-equiv="refresh" content="1; url=/nojavascript/"> Absolute Value Linear Equations and Inequalities | CK-12 Foundation
You are reading an older version of this FlexBook® textbook: CK-12 Algebra I - Honors Go to the latest version.

# 2.7: Absolute Value Linear Equations and Inequalities

Created by: CK-12

## An Explanation and Examples of Absolute Value

Objectives

The lesson objectives for Solving Equations and Inequalities with Absolute Value are:

• An explanation and examples of absolute value
• Solving an absolute value linear equation
• Solving an absolute value linear inequality
• Graphing the solution of an absolute value inequality on a real number line

Introduction

This final section of the chapter introduces you to absolute value. If you remember from the previous chapter, the set of real numbers $(R)$ contain both rational $(Q)$ and irrational $(\overline{Q})$ numbers. The diagram below shows a representation of the set of real numbers.

In this lesson, you will begin the study of absolute value. Absolute value in the real number system is the distance from zero on the number line. It is always a positive number.

As well, absolute value is always written as $|x|$. Using this notation can be translated as “the positive value of $x$”.

Watch This

Guidance

What is the absolute value represented by the number line below?

Remember the definition of absolute value in the real number system is the distance from zero on the number line. Let’s look at the distance from the -6 to zero.

Therefore $|-6|=6$

Example A

$|-7|=?$

$|-7|=7$

Example B

$|7-15|&=?\\|-8|&=?$

$|-8|=8$

Example C

$|-9|-|-2| &= ?\\|-9|-|-2| &= 9-2\\|-9|-|-2| &= 7$

Vocabulary

Absolute Value
Absolute value in the real number system is the distance from zero on the number line. It is always a positive number and is represented using the symbol $|x|$.

Guided Practice

1. $|-50|=?$

2. $\big |- \frac{4}{5} \big |=?$

3. $|-3|-|-4|=?$

Answers

1. $|-50|=?$

$|-50|=50$

2. $\big | -\frac{4}{5} \big |=?$

$\big | - \frac{4}{5}\big |=\frac{4}{5}$

4. $|-3|-|-4| &=?\\|-3|-|-4| &=3-4\\|-3|-|-4| &= -1$

Summary

In this lesson you were introduced to absolute value. Absolute value in the real number system is the distance from zero on the number line. It is always a positive number and is represented using the symbol $|x|$.

Problem Set

Evaluate each of the following using a number line:

1. $|-4.5|=?$
2. $|-1|=?$
3. $\big | -\frac{1}{3} \big |=?$
4. $|4|=?$
5. $|-2|=?$

Evaluate each of the following:

1. $|-1|+|3|=?$
2. $|5|-|-2|=?$
3. $\big | -\frac{1}{2}\big |+ \big | - \frac{2}{3} \big |=?$
4. $|-2.4|-|-1.6|=?$
5. $|-3|-|-2.4|=?$

Evaluate each of the following:

1. $|2-4|=?$
2. $|5-6|=?$
3. $\big | -\frac{1}{2} - \frac{5}{6}\big |=?$
4. $|2.3-3.7|=?$
5. $|7.8 - 9.4|=?$

Answers

Evaluate each of the following using a number line...

1. $|-4.5|=?$ $|-4.5|=4.5$
1. $\big | -\frac{1}{3}\big |=?$ $\big | - \frac{1}{3}\big |=\frac{1}{3}$
1. $|-2|=?$ $|-2|=2$

Evaluate each of the following...

1. $|-1|+|3|=?$

$|-1|+|3|&=1+3\\|-1|+|3|&=4$

1. $\big |-\frac{1}{2}\big |+\big |-\frac{2}{3}\big |=?$

$\bigg |-\frac{1}{2} \bigg |+\bigg |-\frac{2}{3} \bigg |&=\frac{1}{2}+\frac{2}{3}\\\bigg |-\frac{1}{2}\bigg |+\bigg |-\frac{2}{3}\bigg |&=\frac{1}{2}\left({\color{red}\frac{3}{3}}\right)+\frac{2}{3}\left({\color{red}\frac{2}{2}}\right)\\\bigg |-\frac{1}{2}\bigg |+\bigg |-\frac{2}{3}\bigg |&=\frac{3}{6}+\frac{4}{6}\\\bigg |-\frac{1}{2}\bigg |+\bigg |-\frac{2}{3}\bigg |&= \frac{7}{6}$

1. $|-3|-|-2.4|=?$

$|-3|-|-2.4|&=3-2.4\\|-3|-|-2.4|&=0.6$

Evaluate each of the following...

1. $|2-4|=?$

$|2-4|&=|-2|\\|2-4|&=2$

1. $\big |-\frac{1}{2}-\frac{5}{6}\big |=?$

$\bigg |-\frac{1}{2}-\frac{5}{6}\bigg | &=\bigg |-\frac{3}{6}-\frac{5}{6}\bigg |\\\bigg |-\frac{1}{2}-\frac{5}{6}\bigg | &=\bigg |-\frac{8}{6}\bigg |\\\bigg |-\frac{1}{2}-\frac{5}{6}\bigg | &= \bigg |-\frac{4}{3}\bigg |\\\bigg |-\frac{1}{2}-\frac{5}{6}\bigg | &= \frac{4}{3}$

1. $|7.8-9.4|=?$

$|7.8-9.4|&=|-1.6|\\|7.8-9.4| &=1.6$

## Solving an Absolute Value Linear Equation

Introduction

Recall that a linear equation relates mathematical expressions with the equal sign. Linear equations, actually, are considered to be true or false statements because there is only one solution. Remember when you did your checks? You found that the solution for the variable in the linear equation was true and therefore you did all of your algebra correct!

To solve an absolute linear equation you have to remember the same rules that you have learned and used throughout this chapter to solve linear equations with one variable. The difference now is that you are dealing with absolute value. Keep in mind that the absolute value is the positive value of $x$.

Watch This

Guidance

Solve for the variable in the expression: $|3x+1|=4$

Because $|3x+1|=4$, the expression $3x + 1$ is equal to 4 or -4.

$3x+1&=4\\3x+1{\color{red}-1}&=4{\color{red}-1} && \approx\text{Subtract 1 from both sides to isolate the variable}\\3x&=3 && \approx \text{Simplify}\\\frac{3x}{{\color{red}3}}&=\frac{3}{{\color{red}3}} && \approx \text{Divide by 3 to solve for the variable}\\x&=1 && \approx \text{Simplify}\\&& OR\\3x+1&=-4\\3x+1{\color{red}-1}&=-4{\color{red}-1} && \approx \text{Subtract 1 from both sides to isolate the variable}\\3x&=-5 && \approx \text{Simplify}\\\frac{3x}{{\color{red}3}}&=\frac{-5}{{\color{red}3}} && \approx \text{Divide by 3 to solve for the variable}\\x&=\frac{-5}{3} && \approx \text{Simplify}$

With absolute value linear equations, there are actually two solutions! Let’s check and see if this is true.

$&|3x+1|=4 && |3x+1|=4\\& \bigg |3\left({\color{red}\frac{-5}{3}}\right)+1 \bigg |=4 && |3({\color{red}1})+1|=4\\& |-5+1|=|-4|=4 && |4|=4$

So the linear equation has two solutions when there is an absolute function.

Example A

$|d+3|=2.1$

$d+3&=2.1\\d+3{\color{red}-3}&=2.1{\color{red}-3} && \approx \text{Subtract 3 from both sides to isolate the variable}\\d&=0.9 && \approx \text{Simplify}\\& OR\\d+3 &=-2.1\\d+3{\color{red}-3}&=-2.1{\color{red}-3} && \approx \text{Subtract 3 from both sides to isolate the variable}\\d&=-5.1 && \approx \text{Simplify}$

Solutions $= 0.9, -5.1$

Example B

$|2(z+4)|=|5|$

$2(z+4)&=5\\2z+8&=5 && \approx \text{Remove brackets}\\2z+8{\color{red}-8}&=5{\color{red}-8} && \approx \text{Subtract 8 from both sides to isolate the variable}\\2z&=-3 && \approx \text{Simplify}\\\frac{2z}{{\color{red}2}}&=\frac{-3}{{\color{red}2}} && \approx \text{Divide by 2 to solve for the variable}\\z&=\frac{-3}{2} && \approx \text{Simplify}\\& OR\\2(z+4)&=-5 \\2z+8&=-5 && \approx \text{Remove brackets}\\2z+8{\color{red}-8}&=-5{\color{red}-8} && \approx \text{Subtract 8 from both sides to isolate the variable}\\2z&=-13 && \approx \text{Simplify}\\\frac{2z}{{\color{red}2}}&=\frac{-13}{{\color{red}2}} && \approx \text{Divide by 2 to solve for the variable}\\z&=\frac{-13}{2} && \approx \text{Simplify}$

Solutions $= \frac{-3}{2},\frac{-13}{2}$

Example C

$\big|\frac{1}{2}x+3\big|=\big|\frac{4}{5}\big|$

$\frac{1}{2}x+3&=\frac{4}{5}\\\left({\color{red}\frac{5}{5}}\right)\frac{1}{2}x+\left({\color{red}\frac{10}{10}}\right)3&=\left({\color{red}\frac{2}{2}}\right)\frac{4}{5} && \approx \text{Multiply to get common denominator (LCD} = 10)\\\frac{5}{10}x+\frac{30}{10}&=\frac{8}{10} && \approx \text{Simplify}\\5x+30 &= 8 && \approx \text{Simplify}\\5x+30{\color{red}-30}&=8{\color{red}-30} && \approx \text{Subtract 30 from both sides to isolate the variable}\\5x&=-22 && \approx \text{Simplify}\\\frac{5x}{{\color{red}5}}&=\frac{-22}{{\color{red}5}} && \approx \text{Divide by 5 to solve for the variable}\\x &= \frac{-22}{5} && \approx \text{Simplify}\\& OR \\\frac{1}{2}x+3&=\frac{-4}{5}\\\left({\color{red}\frac{5}{5}}\right)\frac{1}{2}x +\left({\color{red}\frac{10}{10}}\right)3&=\left({\color{red}\frac{2}{2}}\right)\frac{-4}{5} && \approx \text{Multiply to get common denominator (LCD} = 10) \\\frac{5}{10}x+\frac{30}{10}&=\frac{-8}{10} && \approx \text{Simplify}\\5x+30 &= -8 && \approx \text{Simplify}\\5x+30{\color{red}-30}&=-8{\color{red}-30} && \approx \text{Subtract 30 from both sides to isolate the variable}\\5x&=-38 && \approx \text{Simplify}\\\frac{5x}{{\color{red}5}}&=\frac{-38}{{\color{red}5}} && \approx \text{Divide by 5 to solve for the variable}\\x&=\frac{-38}{5} && \approx \text{Simplify}$

Solutions $= \frac{-22}{5},\frac{-38}{5}$

Vocabulary

Absolute Value
Absolute value in the real number system is the distance from zero on the number line. It is always a positive number and is represented using the symbol $|x|$.
Linear Equation
A linear equation relates mathematical expressions with the equal sign. Linear equations, actually, are considered to be true or false statements because there is only one solution.

Guided Practice

1. $|4a-2|=3$

2. $|2b-8|-3=4$

3. $\big|\frac{1}{2}c-5\big|=3$

Answers

1. $|4a-2|=3$

$4a-2&=3\\4a-2{\color{red}+2}&=3{\color{red}+2} && \approx \text{Add 2 to both sides to isolate the variable}\\4a&=5 && \approx \text{Simplify}\\\frac{4a}{{\color{red}4}}&=\frac{5}{{\color{red}4}} && \approx \text{Divide by 4 to solve for the variable}\\a &= \frac{5}{4}\\& OR\\4a-2&=-3\\4a-2{\color{red}+2}&=-3{\color{red}+2} && \approx \text{Add 2 to both sides to isolate the variable}\\4a&=-1 && \approx \text{Simplify}\\\frac{4a}{{\color{red}4}}&=\frac{-1}{{\color{red}4}} && \approx \text{Divide by 4 to solve for the variable}\\a&=\frac{-1}{4}$

Solutions $= \frac{5}{4},\frac{-1}{4}$

2. $|2b-8|-3=4$

$2b-8-3&=4\\2b-11&=4 && \approx \text{Combine constant terms on left side of equal sign}\\2b-11{\color{red}+11}&=4{\color{red}+11} && \approx \text{Add 11 to both sides to isolate the variable}\\2b&=15 && \approx \text{Simplify}\\\frac{2b}{{\color{red}2}}&=\frac{15}{{\color{red}2}} && \approx \text{Divide by 2 to solve for the variable}\\b &= \frac{15}{2}\\& OR\\2b-8-3 &=-4\\2b-11 &=-4 && \approx \text{Combine constant terms on left side of equal sign}\\2b-11{\color{red}+11}&=-4{\color{red}+11} && \approx \text{Add 11 to both sides to isolate the variable}\\2b&=7 && \approx \text{Simplify}\\\frac{2b}{{\color{red}2}}&=\frac{7}{{\color{red}2}} && \approx \text{Divide by 2 to solve for the variable}\\b&=\frac{7}{2}$

Solutions $= \frac{15}{2},\frac{7}{2}$

3. $\big| \frac{1}{2}c-5\big|=3$

$\frac{1}{2}c-5 &= 3\\\frac{1}{2}c-\left({\color{red}\frac{2}{2}}\right)5&=\left({\color{red}\frac{2}{2}}\right)3 && \approx \text{Multiply to get common denominator. (LCD} = 2)\\\frac{c}{2}-\frac{10}{2}&=\frac{6}{2} && \approx \text{Simplify}\\c-10 &=6 && \approx \text{Simplify}\\c-10{\color{red}+10}&=6{\color{red}+10} && \approx \text{Add 10 to both sides to isolate the variable}\\c &= 16\\& OR\\\frac{1}{2}c-5&=-3\\\frac{1}{2}c-\left({\color{red}\frac{2}{2}}\right)5&=\left({\color{red}\frac{2}{2}}\right)-3 && \approx \text{Multiply to get common denominator. (LCD} = 2)\\\frac{c}{2}-\frac{10}{2}&=\frac{-6}{2} && \approx \text{Simplify}\\c-10&=-6 && \approx \text{Simplify} \\c-10{\color{red}+10}&=-6{\color{red}+10} && \approx \text{Add 10 to both sides to isolate the variable}\\c&=4$

Solutions $= 16, 4$

Summary

Linear equations in one variable have the form $ax + b = c$, where $a \ne 0$. If you add absolute value to linear equations, the general form of the equation becomes $|ax+b|=c$. The rules for solving linear equations involving absolute values are the same as solving linear equations with absolute values. There is one distinct difference though.

With absolute value equations, there are two solutions. In other words, in order to solve for $x$, you need to do two calculations, namely:

$ax+b=c \ and \ ax+b=-c$

It is also important to note that $c$ must be greater than or equal to 0. In other words, $c \ge 0$. If $c$ is negative, there is no solution. Think about it. Look at the statement below.

$|x|=-1$

The absolute value of $x$ can never be equal to a negative number. Therefore if an absolute value equation is equal to a negative number, there is no solution.

Problem Set

Find the solutions for the variable in each of the following absolute value linear equations.

1. $|t+2|=4$
2. $|r-2|=7$
3. $|5-k|=6$
4. $|6-y|=12$
5. $-6=|1-b|$

Find the solutions for the variable in each of the following absolute value linear equations.

1. $\big|\frac{1}{5}x-3\big|=1$
2. $\big|\frac{1}{2}(r-3)\big|=2$
3. $\big|\frac{1}{3}(f+1)\big|=5$
4. $|3d-11|=-2$
5. $|5w+9|-6=68$

Find the solutions for the variable in each of the following absolute value linear equations.

1. $|5(2t+5)+3(t-1)|=-3$
2. $|2.24x-24.63|=2.25$
3. $|6(5j-3)+2|=14$
4. $|7g-8(g+3)|=1$
5. $|e+4(e+3)|=17$

Answers

Find the solutions for...

1. $|t+2|=4$

$t+2 &= 4\\t+2{\color{red}-2}&=4{\color{red}-2} && \approx \text{Subtract 2 from both sides to isolate the variable}\\t &= 2\\& OR\\t+2 &=-4 && \approx \text{Subtract 2 from both sides to isolate the variable}\\t+2{\color{red}-2}&=-4{\color{red}-2}\\t &= -6$

Solutions $= 2, -6$

1. $|5-k|=6$

$5-k &= 6\\5{\color{red}-5}-k &=6{\color{red}-5} && \approx \text{Subtract 5 from each side to isolate the variable}\\-k &= 1 && \approx \text{Simpliify}\\\frac{-k}{{\color{red}-1}} &= \frac{1}{{\color{red}-1}} && \approx \text{Divide by -1 to solve for the variable}\\k &=-1\\& OR\\5-k &= -6\\5{\color{red}-5}-k &= -6{\color{red}-5}&& \approx \text{Subtract 5 from each side to isolate the variable}\\-k &= -11 && \approx \text{Simpliify}\\\frac{-k}{{\color{red}-1}} &= \frac{-11}{{\color{red}-1}} && \approx \text{Divide by -1 to solve for the variable}\\k &= 11$

Solutions $= -1,11$

1. $-6 = |1-b|$ Since $c < 0$, (for the absolute equation $|ax+b|=c$), there is no solution.

Find the solutions for...

1. $\big|\frac{1}{5}x-3\big|=1$

$\frac{1}{5}x-3&=1\\\frac{1}{5}x-\left({\color{red}\frac{5}{5}}\right)3&=\left({\color{red}\frac{5}{5}}\right)1 && \approx \text{Multiply to get common denominator. (LCD} = 5)\\\frac{x}{5}-\frac{15}{5}&=\frac{5}{5} && \approx \text{Simplify}\\x-15&=5 && \approx \text{Simplify}\\x-15{\color{red}+15}&=5{\color{red}+15} && \approx \text{Add 15 to both sides to isolate the variable}\\x &=20\\& OR\\\frac{1}{5}x-3&=-1\\\frac{1}{5}x-\left({\color{red}\frac{5}{5}}\right)3&=\left({\color{red}\frac{5}{5}}\right)(-1) && \approx \text{Multiply to get common denominator. (LCD} = 5)\\\frac{x}{5}-\frac{15}{5}&=\frac{-5}{5} && \approx \text{Simplify}\\x-15&=-5 && \approx \text{Simplify}\\x-15{\color{red}+15}&=-5{\color{red}+15} && \approx \text{Add 15 to both sides to isolate the variable}\\x &=10$

Solutions $= 20, 10$

1. $\big|\frac{1}{3}(f+1)\big|=5$

$\frac{1}{3}(f+1) &= 5\\\frac{1}{3}f+\frac{1}{3} &= 5 && \approx \text{Remove brackets}\\\frac{1}{3}f+\frac{1}{3}&=\left({\color{red}\frac{3}{3}}\right)5 && \approx \text{Multiply to get common denominator. (LCD} = 3)\\\frac{f}{3}+\frac{1}{3}&=\frac{15}{3} && \approx \text{Simplify}\\f+1 &= 15 && \approx \text{Simplify}\\f+1{\color{red}-1}&=15{\color{red}-1} && \approx \text{Subtract 1 from both sides to isolate and solve for the variable}\\f &=14\\& OR\\\frac{1}{3}(f+1) &= -5\\\frac{1}{3}f+\frac{1}{3} &= -5 && \approx \text{Remove brackets}\\\frac{1}{3}f+\frac{1}{3}&=\left({\color{red}\frac{3}{3}}\right)(-5) && \approx \text{Multiply to get common denominator. (LCD} = 3)\\\frac{f}{3}+\frac{1}{3}&=\frac{-15}{3} && \approx \text{Simplify}\\f+1 &= -15 && \approx \text{Simplify}\\f+1{\color{red}-1}&=-15{\color{red}-1} && \approx \text{Subtract 1 from both sides to isolate and solve for the variable}\\f &=-16$

Solutions $= 14, -16$

1. $|5w+9|-6=68$

$5w+9-6 &=68\\5w+3 &= 68 && \approx \text{Combine constant terms on left side of equal sign}\\5w+3{\color{red}-3}&=68{\color{red}-3} && \approx \text{Subtract 3 from both sides to isolate the variable}\\5w &= 65 && \approx \text{Simplify}\\\frac{5w}{{\color{red}5}}&= \frac{65}{{\color{red}5}} && \approx \text{Divide by 5 to solve for the variable}\\w &= 13 \\& OR \\5w+9-6 &=-68\\5w+3 &= -68 && \approx \text{Combine constant terms on left side of equal sign}\\5w+3{\color{red}-3}&=-68{\color{red}-3} && \approx \text{Subtract 3 from both sides to isolate the variable}\\5w &= -71 && \approx \text{Simplify}\\\frac{5w}{{\color{red}5}}&= \frac{-71}{{\color{red}5}} && \approx \text{Divide by 5 to solve for the variable}\\w &= \frac{-71}{5}$

Solutions $= 13, \frac{-71}{5}$

Find the solutions for...

1. $|5(2t+5)+3(t-1)|=-3$ Since $c < 0$, (for the absolute equation $|ax+b|=c$), there is no solution.
1. $|6(5j-3)+2|=14$

$6(5j-3)+2 &= 14\\30j-18+2 &=14 && \approx \text{Remove brackets}\\30j-16 &= 14 && \approx \text{Combine constant terms on left side of equal sign}\\30j-16{\color{red}+16}&=14{\color{red}+16} && \approx \text{Add 16 to both sides to isolate the variable}\\30j &=30 && \approx \text{Simplify}\\\frac{30j}{{\color{red}30}}&=\frac{30}{\color{red}30} && \text{Divide by 30 to solve for the variable}\\j &=1\\& OR\\6(5j-3)+2 &= -14\\30j-18+2 &=-14 && \approx \text{Remove brackets}\\30j-16 &= -14 && \approx \text{Combine constant terms on left side of equal sign}\\30j-16{\color{red}+16}&=-14{\color{red}+16} && \approx \text{Add 16 to both sides to isolate the variable}\\30j &=2 && \approx \text{Simplify}\\\frac{30j}{{\color{red}30}}&=\frac{2}{\color{red}30} && \approx \text{Divide by 30 to solve for the variable}\\j &=\frac{1}{15}$

Solutions $= 1, \frac{1}{15}$

1. $|e+4(e+3)|=17$

$e+4(e+3)&=17\\e+4e+12 &=17 && \approx \text{Remove brackets}\\5e+12 &=17 && \approx \text{Combine like terms on left side of equal sign}\\5e+12{\color{red}-12} &= 17 {\color{red}-12} && \approx \text{Subtract 12 from sides to isolate the variable}\\5e &= 5 && \approx \text{Simplify}\\\frac{5e}{{\color{red}5}}&=\frac{5}{{\color{red}5}} && \approx \text{Divide by 5 to solve for the variable}\\e &=1\\& OR\\e+4(e+3)&=-17\\e+4e+12 &=-17 && \approx \text{Remove brackets}\\5e+12 &=-17 && \approx \text{Combine like terms on left side of equal sign}\\5e+12{\color{red}-12} &= -17 {\color{red}-12} && \approx \text{Subtract 12 from sides to isolate the variable}\\5e &= -29 && \approx \text{Simplify}\\\frac{5e}{{\color{red}5}}&=\frac{-29}{{\color{red}5}} && \approx \text{Divide by 5 to solve for the variable}\\e &=\frac{-29}{5}$

Solutions $= 1, \frac{-29}{5}$

## Solving an Absolute Value Inequality

Introduction

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

You have just learned that when solving absolute value linear equations, you have to solve for the two related equations. Remember that for $|ax+b|=c$, you had to solve for $ax+b=c$ and $ax+b=-c$. 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 $|ax+b|>c$ $|ax+b| $|ax+b|\ge c$ $|ax+b|\le c$
Equation 1 $ax+b>c$ $ax+b $ax+b \ge c$ $ax+b \le c$
Equation 2 $ax+b<-c$ $ax+b>-c$ $ax+b \le -c$ $ax+b \ge -c$

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

Watch This

Guidance

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 $s =|-32t+150|$, where $s$ 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.

$86 > |-32t+150|$

$86 & > -32t+150\\86 {\color{red}-150} &> -32t+150{\color{red}-150} && \approx \text{Subtract 150 from sides to isolate the variable}\\-64 & > -32t && \approx \text{Simplify}\\\frac{-64}{{\color{red}-32}} & < \frac{-32t}{{\color{red}-32}} && \approx \text{Divide by -32 to solve for the variable. Remember when}\\& && \quad \ \text{dividing by a negative number to reverse the sign of the inequality.}\\t &> 2\\& OR\\-86 & < -32t+150\\-86 {\color{red}-150} & < -32t+150{\color{red}-150} && \approx \text{Subtract 150 from sides to isolate the variable}\\-236 & < -32t && \approx \text{Simplify}\\\frac{-236}{{\color{red}-32}} & > \frac{-32t}{{\color{red}-32}} && \approx \text{Divide by -32 to solve for the variable. Remember when}\\& && \quad \ \text{dividing by a negative number to reverse the sign of the inequality.}\\t &< 7.4$

Therefore when $t >2$ or $t <7.4$, the speed is greater than 64 ft/sec.

Example A

Solve for the absolute value inequality $|g+5|<3$.

$|g+5|<3$

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

Solutions $g <-2$ and $g>-8$.

Example B

Solve for the absolute value inequality $\big|j-\frac{1}{2}\big|>2$.

$\big|j-\frac{1}{2}\big|>2$

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

Solutions $j > \frac{3}{2}$ and $j < \frac{-1}{2}$.

Example C

Solve for the absolute value inequality $|t+1|-3 \ge 2$.

$|t+1|-3 \ge 2$

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

Solutions $t \ge 4$ and $t<0$.

Vocabulary

Absolute Value Linear Inequality
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 can have one of four forms: $ax + b > c, ax + b < c, ax + b \ge c$, or $ax + b \le c$. 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 $|x-1| \ge 9$.

2. Solve for the solution set to the inequality $|-2w+7|<23$.

3. Solve for the solution set to the inequality $|-4+2b|+3 \le 21$.

Answers

1. $|x-1| \ge 9$

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

Therefore solution set is $x \ge 10,x \le -8$.

2. $|-2w+7| < 23$

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

Therefore solution set is $w >-8,w<15$.

3. $|-4+2b|+3 \le 21$

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

Therefore solution set is $b \le 11,b \ge -10$.

Summary

In this lesson you solved absolute value linear inequalities. You found that you solved the linear inequalities using the same rules as you learned in previous lessons. The difference between solving linear inequalities and solving absolute value linear inequalities is that for the absolute value linear inequalities you must solve for the two related inequalities.

If you recall the table from the beginning of this lesson, you remember the absolute value linear inequalities and their related inequalities. So, for example, if you had the absolute value linear inequality $|ax+b| \ge c$, you would need to solve for both $ax+b \ge c$ and $ax+b \le -c$ to find the solution set.

It is also important to remember, as you found with absolute value linear equations, that if $c < 0$, there is no solution. If you think about it with an example it may seem clearer. Look at the example $|x + 3| > -5$. As you found in the previous lesson, there is no absolute value of $x + 3$ that will give you a negative number as an answer.

Problem Set

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

1. $|p-16|>10$
2. $|r+2|<5$
3. $|3-2k|\ge -1$
4. $|8-y|>5$
5. $8 \ge |5d-2|$

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

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

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

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

Answers

Find the solution sets for...

1. $|p-16|>10$

$p-16 & > 10\\p-16{\color{red}+16}&>10{\color{red}+16} && \approx \text{Add 16 to sides to isolate and solve for the variable}\\p &> 26\\& OR\\p-16 & < -10\\p-16{\color{red}+16}&<-10{\color{red}+16} && \approx \text{Add 16 to sides to isolate and solve for the variable}\\p &< 6$

The solution sets are $p>26,p<6$.

1. $|3-2k|\ge -1$ Since $c < 0$, (for the absolute value inequality $|ax+b| \ge c$), there is no solution.
1. $8 \ge |5d-2|$

$8 &\ge 5d-2\\8{\color{red}+2} &\ge 5d-2{\color{red}+2} && \approx \text{Add 2 to each side to isolate the variable}\\10 &\ge 5d && \approx \text{Simplify}\\\frac{5d}{{\color{red}5}} & \le \frac{10}{{\color{red}5}} && \approx \text{Divide by 5 to solve for the variable}\\d &\le 2\\& OR\\-8 &\le 5d-2\\-8{\color{red}+2} &\le 5d-2{\color{red}+2} && \approx \text{Add 2 to each side to isolate the variable}\\-6 &\le 5d && \approx \text{Simplify}\\\frac{5d}{{\color{red}5}} & \ge \frac{-6}{{\color{red}5}} && \approx \text{Divide by 5 to solve for the variable}\\d &\ge \frac{-6}{5}$

The solution sets are $d \le 2, d \ge \frac{-6}{5}$.

Find the solution sets for...

1. $|s+2|-5>8$

$s+2-5 & >8\\s-3 &> 8 && \approx \text{Combine constant terms on left side of inequality}\\s-3{\color{red}+3} &> 8{\color{red}+3} && \approx \text{Add 3 to both sides to isolate and solve for the variable}\\s & > 11\\& OR\\s+2-5 & <-8\\s-3 &< -8 && \approx \text{Combine constant terms on left side of inequality}\\s-3{\color{red}+3} &< -8{\color{red}+3} && \approx \text{Add 3 to both sides to isolate and solve for the variable}\\s & <-5$

The solution sets are $s>11,s<-5$.

1. $|2q+1|-5 \le 7$

$2q+1-5 &\le 7\\2q-4 &\le 7 && \approx \text{Combine constant terms on left side of inequality}\\2q-4{\color{red}+4} &\le 7{\color{red}+4} && \approx \text{Add 4 to both sides to isolate the variable}\\2q &\le 11 && \approx \text{Simplify}\\\frac{2q}{{\color{red}2}} & \le \frac{11}{{\color{red}2}} && \approx \text{Divide by 2 to solve for the variable}\\q &\le \frac{11}{2}\\& OR\\2q+1-5 &\ge -7\\2q-4 &\ge -7 && \approx \text{Combine constant terms on left side of inequality}\\2q-4{\color{red}+4} &\ge -7{\color{red}+4} && \approx \text{Add 4 to both sides to isolate the variable}\\2q &\ge -3 && \approx \text{Simplify}\\\frac{2q}{{\color{red}2}} & \ge \frac{-3}{{\color{red}2}} && \approx \text{Divide by 2 to solve for the variable}\\q &\ge \frac{-3}{2}$

The solution sets are $q \le \frac{11}{2},q \ge \frac{-3}{2}$.

1. $|-2(e+4)|>17$

$-2(e+4) &>17\\-2e-8 &>17 && \approx \text{Remove brackets}\\-2e-8{\color{red}+8}& > 17{\color{red}+8} && \approx \text{Add 8 to both sides to isolate the variable}\\-2e &>25 && \approx \text{Simplify}\\\frac{-2e}{{\color{red}-2}} &< \frac{25}{{\color{red}-2}} && \approx \text{Divide by -2 to solve for the variable. Reverse sign of the inequality.}\\e & < \frac{-25}{2}\\& OR\\-2(e+4) &<-17\\-2e-8 &<-17 && \approx \text{Remove brackets}\\-2e-8{\color{red}+8}& < -17{\color{red}+8} && \approx \text{Add 8 to both sides to isolate the variable}\\-2e &< 9 && \approx \text{Simplify}\\\frac{-2e}{{\color{red}-2}} &> \frac{9}{{\color{red}-2}} && \approx \text{Divide by -2 to solve for the variable. Reverse sign of the inequality.}\\e & > \frac{-9}{2}$

The solution sets are $e < \frac{-25}{2},e > \frac{-9}{2}$.

Find the solution sets for...

1. $|-5x-3(2x-1)|>3$

$-5x-3(2x-1)&>3\\-5x-6x+3 &> 3 && \approx \text{Remove brackets}\\-11x+3 &> 3 && \approx \text{Combine like terms on left side of inequality}\\-11x+3{\color{red}-3}&>3{\color{red}-3} && \approx \text{Subtract 11 from both sides to isolate the variable}\\-11x & > 0 && \approx \text{Simplify}\\\frac{-11x}{{\color{red}-11}} &< \frac{0}{{\color{red}-11}} && \approx \text{Divide by -11 to solve for the variable. Reverse sign of the inequality.}\\x &< 0\\& OR \\-5x-3(2x-1)&<-3\\-5x-6x+3 &< -3 && \approx \text{Remove brackets}\\-11x+3 &< -3 && \approx \text{Combine like terms on left side of inequality}\\-11x+3{\color{red}-3}&<-3{\color{red}-3} && \approx \text{Subtract 11 from both sides to isolate the variable}\\-11x & <-6 && \approx \text{Simplify}\\\frac{-11x}{{\color{red}-11}} &> \frac{-6}{{\color{red}-11}} && \approx \text{Divide by -11 to solve for the variable. Reverse sign of the inequality.}\\x &> \frac{6}{11}$

The solution sets are $x<0,x>\frac{6}{11}$.

1. $|-2(r+3.1)| \le -1.4$ Since $c < 0$, (for the absolute value inequality $|ax+b|\le c$), there is no solution.
1. $\big|-2\left(e-\frac{3}{4}\right)\big| \ge 3$

$-2\left(e-\frac{3}{4}\right) &\ge 3\\-2e+\frac{6}{4} &\ge 3 && \approx \text{Remove brackets}\\\left({\color{red}\frac{4}{4}}\right)(-2e)+\frac{6}{4} & \ge \left({\color{red}\frac{4}{4}}\right)3 && \approx \text{Multiply to get common denominator. (LCD} = 4)\\\frac{-8e}{4}+\frac{6}{4} &\ge \frac{12}{4} && \approx \text{Simplify}\\-8e+6 &\ge 12 && \approx \text{Simplify}\\-8e+6{\color{red}-6} & \ge 12{\color{red}-6} && \approx \text{Subtract 6 from sides to isolate the variable}\\-8e &\ge 6 && \approx \text{Simplify}\\\frac{-8e}{{\color{red}-8}} &\le \frac{6}{{\color{red}-8}} && \approx \text{Divide by -8 to solve for the variable. Reverse sign of inequality.}\\e & \le \frac{-3}{4}\\& OR \\-2\left(e-\frac{3}{4}\right) &\le -3\\-2e+\frac{6}{4} &\le -3 && \approx \text{Remove brackets}\\\left({\color{red}\frac{4}{4}}\right)(-2e)+\frac{6}{4} & \le \left({\color{red}\frac{4}{4}}\right)(-3) && \approx \text{Multiply to get common denominator. (LCD} = 4)\\\frac{-8e}{4}+\frac{6}{4} &\le \frac{-12}{4} && \approx \text{Simplify}\\-8e+6 &\le -12 && \approx \text{Simplify}\\-8e+6{\color{red}-6} & \le -12{\color{red}-6} && \approx \text{Subtract 6 from sides to isolate the variable}\\-8e &\le -18 && \approx \text{Simplify}\\\frac{-8e}{{\color{red}-8}} &\ge \frac{-18}{{\color{red}-8}} && \approx \text{Divide by -8 to solve for the variable. Reverse sign of inequality.}\\e & \ge \frac{9}{4}$

The solution sets are $e \le \frac{-3}{4}, e \ge \frac{9}{4}$.

## Graphing the Solution of an Absolute Value Inequality on a Real Number Line

Introduction

In this last lesson of Chapter - Can You Make It True, you will learn to graph the solution of an absolute value inequality on a number line. In a previous lesson, you graphed linear inequalities on number lines. An example can be seen below.

For $x > 5$, the graph can be shown as:

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? As you learned in the previous lesson, with absolute value linear inequalities, there are two solution sets. Therefore there would be two sections of the number line showing solutions.

For $|t|>5$, you would actually solve for $t > 5$ and $t <-5$. If you were to graph this solution on a number line it would look like the following:

Watch This

Guidance

Solve the following inequality and graph the solution on a number line.

$|x+2|\le 3$

First solve the inequality:

$x+2 &\le 3\\x+2{\color{red}-2} &\le 3{\color{red}-2} && \approx \text{Subtract 2 from both sides to isolate the variable}\\x &\le 1 && \approx \text{Simplify}\\& OR\\x+2 &\ge -3\\x+2{\color{red}-2} &\ge -3{\color{red}-2} && \approx \text{Subtract 2 from both sides to isolate the variable}\\x &\ge -5 && \approx \text{Simplify}$

The solution sets are $x \le 1,x \ge -5$.

Representing on a number line

Example A

Represent the solution set to the following inequality on a number line.

$|2x| &\ge 6\\2x & \ge 6\\\frac{2x}{{\color{red}2}} & \ge \frac{6}{{\color{red}2}} && (\approx \text{Divide by 2 to isolate and solve for the variable})\\x & \ge 3 && (\approx \text{Simplify})\\& OR\\2x & \le -6\\\frac{2x}{{\color{red}2}} & \le \frac{-6}{{\color{red}2}} && (\approx \text{Divide by 2 to isolate and solve for the variable})\\x & \le -3 && (\approx \text{Simplify})$

The solution sets are $x \ge 3, x \le -3$.

Example B

Solve the following inequality and graph the solution on a number line.

$|x+1| &> 3 && (\approx \text{Divide both sides by 2 to solve for the variable})\\x+1 &>3\\x+1{\color{red}-1} &> 3{\color{red}-1} && (\approx \text{Subtract 1 from both sides of the inequality sign})\\x & > 2\\& OR\\x+1 &< -3\\x+1{\color{red}-1} &< -3{\color{red}-1} && (\approx \text{Subtract 1 from both sides of the inequality sign})\\x & < -4$

The solution sets are $x>2,x<-4$.

Example C

Solve the following inequality and graph the solution on a number line.

$\bigg |x-\frac{5}{2} \bigg | &< 1\\x-\frac{5}{2} &< 1\\\left({\color{red}\frac{2}{2}}\right)x-\frac{5}{2} &< \left(\frac{{\color{red}2}}{{\color{red}2}}\right)1\\\frac{2x}{2}-\frac{5}{2}&<\frac{2}{2}\\2x-5&<2 && (\approx \text{Simplify})\\2x-5{\color{red}+5} &< 2{\color{red}+5} && (\approx \text{Subtract 5 to isolate the variable})\\2x &< 7&& (\approx \text{Simplify})\\\frac{2x}{{\color{red}2}} &< \frac{7}{{\color{red}2}}\\x &< \frac{7}{2}\\& OR\\x-\frac{5}{2} &> -1\\\left({\color{red}\frac{2}{2}}\right)x-\frac{5}{2} &> \left(\frac{{\color{red}2}}{{\color{red}2}}\right)(-1) && (\approx \text{Multiply to get common denominator (LCD} = 2))\\\frac{2x}{2}-\frac{5}{2}&<\frac{-2}{2}&& (\approx \text{Simplify})\\2x-5&>-2 && (\approx \text{Simplify})\\2x-5{\color{red}+5} &> -2{\color{red}+5} && (\approx \text{Subtract 5 to isolate the variable})\\2x &> 3&& (\approx \text{Simplify})\\\frac{2x}{{\color{red}2}} &> \frac{3}{{\color{red}2}} && (\approx \text{Divide both sides by 2 to solve for the variable})\\x &> \frac{3}{2}$

The solution sets are $x<\frac{7}{2}, x >\frac{3}{2}$.

Vocabulary

Absolute Value Linear Inequality
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 $ax + b > -c$.
Number Line
A number line is a line that matches a set of points and a set of numbers one to one.

It is often used in mathematics to show mathematical computations.

Guided Practice

1. Represent the solution set to the inequality $|2x+3|<6$ on a number line.

2. Represent the solution set to the inequality $|32x-16| \ge 32$ on a number line.

3. Represent the solution set to the inequality $|x-21.5|>12.5$ on a number line.

Answers

1. $|2x+3| >5$

$2x+3&>5\\2x+3{\color{red}-3}&>5{\color{red}-3} && (\approx \text{Subtract 3 from both sides of the inequality sign})\\2x &> 2 && (\approx \text{Simplify})\\\frac{2x}{{\color{red}2}}&>\frac{2}{{\color{red}2}} && (\approx \text{Divide by 2 to solve for the variable})\\x &>1\\& OR\\2x+3&<-5\\2x+3{\color{red}-3}&<-5{\color{red}-3} && (\approx \text{Subtract 3 from both sides of the inequality sign})\\2x &< -8 && (\approx \text{Simplify})\\\frac{2x}{{\color{red}2}}&>\frac{-8}{{\color{red}2}} && (\approx \text{Divide by 2 to solve for the variable})\\x &<-4$

The solution sets are $x>1$ and $x<-4$.

2. $|32x-16| \ge 32$

$32x-16 &\ge 32\\32x-16{\color{red}+16}&\ge 32{\color{red}+16} && (\approx\text{Add 16 to both sides of the inequality sign})\\32x &\ge 48&& (\approx\text{Simplify})\\\frac{32x}{{\color{red}32}} &\ge \frac{48}{{\color{red}32}}&& (\approx \text{Divide by 32 to solve for the variable})\\x &\ge \frac{3}{2}\\& OR \\32x-16 &\le -32\\32x-16{\color{red}+16}&\le -32{\color{red}+16} && (\approx\text{Add 16 to both sides of the inequality sign})\\32x &\le -16&& (\approx\text{Simplify})\\\frac{32x}{{\color{red}32}} &\le \frac{-16}{{\color{red}32}}&& (\approx \text{Divide by 32 to solve for the variable})\\x &\le -\frac{1}{2}$

The solution sets are $x \ge \frac{3}{2}, x \le -\frac{1}{2}$.

3. $|x-21.5|>12.5$

$x-21.5 &> 12.5\\x-21.5{\color{red}+21.5}&>12.5{\color{red}+21.5} && (\approx \text{Add 21.5 to both sides to isolate the variable})\\x &>34 &&(\approx \text{Simplify})\\& OR\\x-21.5 &< -12.5\\x-21.5{\color{red}+21.5}&<-12.5{\color{red}+21.5} && (\approx \text{Add 21.5 to both sides to isolate the variable})\\x &<9 &&(\approx \text{Simplify})$

The solution sets are $x<9,x>34$.

Summary

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. In other words, 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 $\le$ and $\ge$.

Problem Set

1. Represent the solution sets to the absolute value inequality $|3-2x|<3$ on a number line.
2. Represent the solution sets to the absolute value inequality $2\big|\frac{2x}{3}+1\big|\ge 4$ on a number line.
3. Represent the solution sets to the absolute value inequality $\big|\frac{2g-9}{4}\big|<1$ on a number line.
4. Represent the solution sets to the absolute value inequality $\big|\frac{4}{3}x-5\big|\ge 7$ on a number line.
5. Represent the solution sets to the absolute value inequality $|2x+5|+4 \ge 1$ on a number line.

Answers

1. $|3-2x| < 3$

$3-2x &<3\\3{\color{red}-3}-2x &< 3{\color{red}-3} && (\approx \text{Subtract 3 from both sides of the inequality sign})\\-2x &< 0 && (\approx \text{Simplify})\\\frac{-2x}{{\color{red}-2}} &< \frac{0}{{\color{red}-2}} && (\approx \text{Divide by -2 to solve for the variable, reverse sign of inequality})\\x &< 0\\& OR\\3-2x &>-3\\3{\color{red}-3}-2x &> -3{\color{red}-3} && (\approx \text{Subtract 3 from both sides of the inequality sign})\\-2x &> -6 && (\approx \text{Simplify})\\\frac{-2x}{{\color{red}-2}} &< \frac{-6}{{\color{red}-2}} && (\approx \text{Divide by -2 to solve for the variable, reverse sign of inequality})\\x &< 3$

The solution sets are $x<0,x<3$.

1. $\big|\frac{2g-9}{4}\big|<1$

$\frac{2g-9}{4} &< 1\\\frac{2g-9}{4}({\color{red}4}) &< 1({\color{red}4}) && (\approx \text{Multiply both sides by 4 to remove the fraction})\\2g-9 &< 4 && (\approx \text{Simplify})\\29-9{\color{red}+9} &< 4{\color{red}+9} && (\approx \text{Add 9 to both sides of the inequality sign})\\2g &< 13 && (\approx \text{Simplify})\\\frac{29}{{\color{red}2}} &< \frac{13}{{\color{red}2}} && (\approx \text{Divide by 2 to solve for the variable})\\g &< \frac{13}{2}\\& OR\\\frac{2g-9}{4} &> -1\\\frac{2g-9}{4}({\color{red}4}) &> -1({\color{red}4}) && (\approx \text{Multiply both sides by 4 to remove the fraction})\\2g-9 &> -4 && (\approx \text{Simplify})\\29-9{\color{red}+9} &> -4{\color{red}+9} && (\approx \text{Add 9 to both sides of the inequality sign})\\2g &> 5 && (\approx \text{Simplify})\\\frac{29}{{\color{red}2}} &> \frac{5}{{\color{red}2}} && (\approx \text{Divide by 2 to solve for the variable})\\g &> \frac{5}{2}$

The solution sets are $g < \frac{13}{2},g > \frac{5}{2}$.

1. $|2x+5|+4 \ge 1$

$2x+5+4 &\ge 1\\2x+9 &\ge 1 && (\approx \text{Combine common constant terms})\\2x+9{\color{red}-9} &\ge 1{\color{red}-9}&& (\approx \text{Subtract 9 from both sides to isolate the variable})\\2x &\ge -8&& (\approx \text{Simplify})\\\frac{2x}{{\color{red}2}} &\ge \frac{-8}{{\color{red}2}}&& (\approx \text{Divide by 2 to solve for the variable})\\x &\ge -4\\& OR\\2x+5+4 &\le -1\\2x+9 &\le -1 && (\approx \text{Combine common constant terms})\\2x+9{\color{red}-9} &\le -1{\color{red}-9}&& (\approx \text{Subtract 9 from both sides to isolate the variable})\\2x &\le -10&& (\approx \text{Simplify})\\\frac{2x}{{\color{red}2}} &\le \frac{-10}{{\color{red}2}}&& (\approx \text{Divide by 2 to solve for the variable})\\x &\le -5$

The solution sets are $x \ge-4,x \le-5$.

## Summary

Throughout this chapter you have been learning about solving linear equations with one variable and solving linear inequalities with one variable. You have learned that whether you have decimals, fractions, brackets, or variables on both sides, the rules stay basically the same for all of these expressions. In other words, the key is to isolate the variable and solve for it. You do this by making sure that whatever you do to one side of the equation you do to the other.

In this final lesson of Chapter 2, you were introduced to absolute value expressions. Absolute value expressions mean that the variable is always in the positive. In fact, if the absolute value linear equation is equal to a negative number or the absolute value linear inequality is greater than, less than, greater than or equal to, or less than or equal to a negative number, the linear equation or the inequality simply have no solution.

In solving absolute value expressions, for both linear equations and linear inequalities, the rules remain the same with one addition. Because you are dealing with the absolute function, you have to solve for the two related expressions when solving a linear equation or inequality.

In the final section of the lesson and chapter, you had the opportunity to solve absolute value linear inequalities and then graph them on a number line. This combines the concepts learned in the previous lesson and those from this lesson as you complete this chapter. Remember that the same rules apply. The open circle indicates the > and< symbol is used in the solution set. The closed circle indicates that the $\le$ and $\ge$ symbols are used in the solution set.

Jan 16, 2013

## Last Modified:

Jan 14, 2015
Files can only be attached to the latest version of None

# Reviews

Please wait...
Please wait...
Image Detail
Sizes: Medium | Original

CK.MAT.ENG.SE.1.Algebra-I---Honors.2.7