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2.13: Solutions to Absolute Value Equations

Difficulty Level: At Grade Created by: CK-12

Solve for the variable in the expression: \begin{align*}|3x+1|=4\end{align*}|3x+1|=4

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Khan Academy Absolute Value Equations

Guidance

Recall that a linear equation relates mathematical expressions with an equals sign. Linear equations, actually, are considered to be true or false statements because there is only one solution.

To solve an absolute linear equation you have to remember the same rules that you have learned and used 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 \begin{align*}x\end{align*}x.

With absolute value equations, there are usually two solutions. In order to solve for \begin{align*}x\end{align*}x, you need to solve two equations, namely:

\begin{align*}ax+b=c \ \text{and} \ ax+b=-c\end{align*}

ax+b=c and ax+b=c

It is also important to note that \begin{align*}c\end{align*}c must be greater than or equal to 0. In other words, \begin{align*}c \ge 0\end{align*}c0. If \begin{align*}c\end{align*}c is negative, there is no solution. Think about it. Look at the statement below.

\begin{align*}|x|=-1\end{align*}

|x|=1

The absolute value of \begin{align*}x\end{align*}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.

Example A

\begin{align*}|d+3|=2.1\end{align*}|d+3|=2.1

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

d+3d+33dd+3d+33d=2.1=2.13=0.9OR=2.1=2.13=5.1 Subtract 3 from both sides to isolate the variableSimplifySubtract 3 from both sides to isolate the variableSimplify

Solutions \begin{align*}= 0.9, -5.1\end{align*}=0.9,5.1

Example B

\begin{align*}|2(z+4)|=|5|\end{align*}|2(z+4)|=|5|

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

2(z+4)2z+82z+882z2z2z2(z+4)2z+82z+882z2z2z=5=5=58=3=32=32OR=5=5=58=13=132=132Remove bracketsSubtract 8 from both sides to isolate the variableSimplifyDivide by 2 to solve for the variableSimplifyRemove bracketsSubtract 8 from both sides to isolate the variableSimplifyDivide by 2 to solve for the variableSimplify

Solutions \begin{align*}= \frac{-3}{2},\frac{-13}{2}\end{align*}=32,132

Example C

\begin{align*}\big|\frac{1}{2}x+3\big|=\big|\frac{4}{5}\big|\end{align*}12x+3=45

\begin{align*}\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} && \text{Multiply to get common denominator (LCD} = 10)\\ \frac{5}{10}x+\frac{30}{10}&=\frac{8}{10} && \text{Simplify}\\ 5x+30 &= 8 && \text{Simplify}\\ 5x+30{\color{red}-30}&=8{\color{red}-30} && \text{Subtract 30 from both sides to isolate the variable}\\ 5x&=-22 && \text{Simplify}\\ \frac{5x}{{\color{red}5}}&=\frac{-22}{{\color{red}5}} && \text{Divide by 5 to solve for the variable}\\ x &= \frac{-22}{5} && \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} && \text{Multiply to get common denominator (LCD} = 10) \\ \frac{5}{10}x+\frac{30}{10}&=\frac{-8}{10} && \text{Simplify}\\ 5x+30 &= -8 && \text{Simplify}\\ 5x+30{\color{red}-30}&=-8{\color{red}-30} && \text{Subtract 30 from both sides to isolate the variable}\\ 5x&=-38 && \text{Simplify}\\ \frac{5x}{{\color{red}5}}&=\frac{-38}{{\color{red}5}} && \text{Divide by 5 to solve for the variable}\\ x&=\frac{-38}{5} && \text{Simplify}\end{align*}

12x+3(55)12x+(1010)3510x+30105x+305x+30305x5x5x12x+3(55)12x+(1010)3510x+30105x+305x+30305x5x5x=45=(22)45=810=8=830=22=225=225OR=45=(22)45=810=8=830=38=385=385Multiply to get common denominator (LCD=10)SimplifySimplifySubtract 30 from both sides to isolate the variableSimplifyDivide by 5 to solve for the variableSimplifyMultiply to get common denominator (LCD=10)SimplifySimplifySubtract 30 from both sides to isolate the variableSimplifyDivide by 5 to solve for the variableSimplify

Solutions \begin{align*}= \frac{-22}{5},\frac{-38}{5}\end{align*}

Concept Problem Revisited

Solve for the variable in the expression: \begin{align*}|3x+1|=4\end{align*}

Because \begin{align*}|3x+1|=4\end{align*}, the expression \begin{align*}3x + 1\end{align*} is equal to 4 or -4.

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

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

\begin{align*}&|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\end{align*}

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

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 \begin{align*}|x|\end{align*}.
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. \begin{align*}|4a-2|=3\end{align*}

2. \begin{align*}|2b-8|-3=4\end{align*}

3. \begin{align*}\big|\frac{1}{2}c-5\big|=3\end{align*}

Answers:

1. \begin{align*}|4a-2|=3\end{align*}

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

Solutions \begin{align*}= \frac{5}{4},\frac{-1}{4}\end{align*}

2. \begin{align*}|2b-8|-3=4\end{align*}

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

Solutions \begin{align*}= \frac{15}{2},\frac{7}{2}\end{align*}

3. \begin{align*}\big| \frac{1}{2}c-5\big|=3\end{align*}

\begin{align*}\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 && \text{Multiply to get common denominator. (LCD} = 2)\\ \frac{c}{2}-\frac{10}{2}&=\frac{6}{2} && \text{Simplify}\\ c-10 &=6 && \text{Simplify}\\ c-10{\color{red}+10}&=6{\color{red}+10} && \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 && \text{Multiply to get common denominator. (LCD} = 2)\\ \frac{c}{2}-\frac{10}{2}&=\frac{-6}{2} && \text{Simplify}\\ c-10&=-6 && \text{Simplify} \\ c-10{\color{red}+10}&=-6{\color{red}+10} && \text{Add 10 to both sides to isolate the variable}\\ c&=4\end{align*}

Solutions \begin{align*}= 16, 4\end{align*}

Practice

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

  1. \begin{align*}|t+2|=4\end{align*}
  2. \begin{align*}|r-2|=7\end{align*}
  3. \begin{align*}|5-k|=6\end{align*}
  4. \begin{align*}|6-y|=12\end{align*}
  5. \begin{align*}-6=|1-b|\end{align*}

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

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

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

  1. \begin{align*}|5(2t+5)+3(t-1)|=-3\end{align*}
  2. \begin{align*}|2.24x-24.63|=2.25\end{align*}
  3. \begin{align*}|6(5j-3)+2|=14\end{align*}
  4. \begin{align*}|7g-8(g+3)|=1\end{align*}
  5. \begin{align*}|e+4(e+3)|=17\end{align*}

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Difficulty Level:

At Grade

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Date Created:

Dec 19, 2012

Last Modified:

Apr 29, 2014
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