Solved Examples: Absolute Value Equations

Samuel Dominic Chukwuemeka (SamDom For Peace) As applicable, verify your answers with the Expressions and Equations Calculators

For ACT Students
The ACT is a timed exam...60 questions for 60 minutes
This implies that you have to solve each question in one minute.
Some questions will typically take less than a minute a solve.
Some questions will typically take more than a minute to solve.
The goal is to maximize your time. You use the time saved on those questions you solved in less than a minute, to solve the questions that will take more than a minute.
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All work is shown to satisfy (and actually exceed) the minimum for awarding method marks.
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For NSC Students
For the Questions:
Any space included in a number indicates a comma used to separate digits...separating multiples of three digits from behind.
Any comma included in a number indicates a decimal point.
For the Solutions:
Decimals are used appropriately rather than commas
Commas are used to separate digits appropriately.

Solve the absolute value equations.
Show all work.
Check all solutions.

(1.) ACT The equation |2x - 8| + 3 = 5 has 2 solutions.
Those solutions are equal to the solutions to which of the following pairs of equations?

$ F.\:\: 2x - 5 = 5 \\[3ex] -2x - 5 = -5 \\[5ex] G.\:\: 2x - 8 = 2 \\[3ex] -2x - 8 = 2 \\[5ex] H.\:\: 2x - 8 = 8 \\[3ex] -(2x - 8) = 8 \\[5ex] J.\:\: 2x - 8 = 2 \\[3ex] -(2x - 8) = 8 \\[5ex] K.\:\: 2x - 8 = 2 \\[3ex] -(2x - 8) = 2 $


$|2x - 8| + 3 = 5$

This means that:

$ |2x - 8| = 5 - 3 \\[3ex] |2x - 8| = 2 \\[3ex] So,\:\: \\[3ex] 2x - 8 = 2 \\[3ex] OR \\[3ex] -(2x - 8) = 2 $
(2.) Solve the following equation for x: |x| = 8


$ |x| = 8 \\[3ex] \implies \\[3ex] x = 8 \;\;\;OR\;\;\; -x = 8 \\[3ex] x = 8 \;\;\;OR\;\;\; x = -8 \\[3ex] $ Check
$x = 8$, $x = -8$
LHS RHS
$ |x| \\[3ex] x = 8 \\[3ex] |8| \\[3ex] 8 $
$ x = -8 \\[3ex] |-8| \\[3ex] 8 $ 		$8$
(3.)

$ (a.)\;\; |-2x + 6| = 6 \\[3ex] (b.)\;\; |-2x - 1| = 11 \\[3ex] (c.)\;\; |-4 + 5x| = 16 \\[3ex] $

$ (a.) \\[3ex] |-2x + 6| = 6 \\[3ex] -2x + 6 = 6 \;\;\;OR\;\;\; -(-2x + 6) = 6 \\[3ex] -2x = 6 - 6 \;\;\;OR\;\;\; 2x - 6 = 6 \\[3ex] -2x = 0 \;\;\;OR\;\;\; 2x = 6 + 6 \\[3ex] x = \dfrac{0}{-2} \;\;\;OR\;\;\; 2x = 12 \\[5ex] x = 0 \;\;\;OR\;\;\; x = \dfrac{12}{2} \\[5ex] x = 0 \;\;\;OR\;\;\; x = 6 \\[3ex] $ Check
$x = 0$, $x = 6$
LHS RHS
$ |-2x + 6| \\[3ex] x = 0 \\[3ex] |-2(0) + 6| \\[3ex] |0 + 6| \\[3ex] |6| \\[3ex] 6 $
$ x = 6 \\[3ex] |-2(6) + 6| \\[3ex] |-12 + 6| \\[3ex] |-6| \\[3ex] 6 $ 		$6$


$ (b.) \\[3ex] |-2x - 1| = 11 \\[3ex] -2x - 1 = 11 \;\;\;OR\;\;\; -(-2x - 1) = 11 \\[3ex] -2x = 11 + 1 \;\;\;OR\;\;\; -2x - 1 = \dfrac{11}{-1} \\[5ex] -2x = 12 \;\;\;OR\;\;\; -2x - 1 = -11 \\[3ex] x = \dfrac{12}{-2} \;\;\;OR\;\;\; -2x = -11 + 1 \\[5ex] x = -6 \;\;\;OR\;\;\; -2x = -10 \\[3ex] x = -6 \;\;\;OR\;\;\; x = \dfrac{-10}{-2} \\[5ex] x = -6 \;\;\;OR\;\;\; x = 5 \\[3ex] $ Check
$x = -6$, $x = 5$
LHS RHS
$ |-2x - 1| \\[3ex] x = -6 \\[3ex] |-2(-6) - 1| \\[3ex] |12 - 1| \\[3ex] |11| \\[3ex] 11 $
$ x = 5 \\[3ex] |-2(5) - 1| \\[3ex] |-10 - 1| \\[3ex] |-11| \\[3ex] 11 $ 		$11$


$ (c.) \\[3ex] |-4 + 5x| = 16 \\[3ex] -4 + 5x = 16 \;\;\;OR\;\;\; -(-4 + 5x) = 16 \\[3ex] 5x = 16 + 4 \;\;\;OR\;\;\; -4 + 5x = -16 \\[3ex] 5x = 20 \;\;\;OR\;\;\; 5x = -16 + 4 \\[3ex] x = \dfrac{20}{5} \;\;\;OR\;\;\; 5x = -12 \\[5ex] x = 4 \;\;\;OR\;\;\; x = -\dfrac{12}{5} \\[5ex] $ Check
$x = 4$, $x = -\dfrac{12}{5}$
LHS RHS
$ |-4 + 5x| \\[3ex] x = 4 \\[3ex] |-4 + 5(4)| \\[3ex] |-4 + 20| \\[3ex] |16| \\[3ex] 16 $
$ x = -\dfrac{12}{5} \\[5ex] \left|-4 + 5\left(-\dfrac{12}{5}\right)\right| \\[5ex] |-4 - 12| \\[3ex] |-16| \\[3ex] 16 $ 		$16$
(4.) $26 - |8x - 3| = 6$


$ 26 - |8x - 3| = 6 \\[3ex] 26 - 6 = |8x - 3| \\[3ex] 20 = |8x - 3| \\[3ex] |8x - 3| = 20 \\[3ex] This\:\: means\:\: that \\[3ex] 8x - 3 = 20 \:\:OR\:\: -(8x - 3) = 20 \\[3ex] 8x - 3 = 20 \\[3ex] 8x = 20 + 3 \\[3ex] 8x = 23 \\[3ex] x = \dfrac{23}{8} \\[5ex] OR \\[3ex] -(8x - 3) = 20 \\[3ex] Divide\:\: both\:\: sides\:\: by\:\: -1 \\[3ex] 8x - 3 = -20 \\[3ex] 8x = -20 + 3 \\[3ex] 8x = -17 \\[3ex] x = -\dfrac{17}{8} \\[5ex] $ Check
$x = \dfrac{23}{8}$, $x = -\dfrac{17}{8}$
LHS RHS
$ 26 - |8x - 3| \\[3ex] x = \dfrac{23}{8} \\[5ex] 26 - \left|8\left(\dfrac{23}{8}\right) - 3\right| \\[5ex] 26 - |23 - 3| \\[3ex] 26 - |20| \\[3ex] 26 - 20 \\[3ex] 6 \\[3ex] $ $x = \dfrac{23}{8}$ is a solution

$ 26 - |8x - 3| \\[3ex] x = -\dfrac{17}{8} \\[5ex] 26 - \left|8\left(-\dfrac{17}{8}\right) - 3\right| \\[5ex] 26 - |-17 - 3| \\[3ex] 26 - |-20| \\[3ex] 26 - 20 \\[3ex] 6 \\[3ex] $ $x = -\dfrac{17}{8}$ is a solution 		$6$
(5.) $|-2p + 2| + 4 = 12$


$ |-2p + 2| + 4 = 12 \\[3ex] |-2p + 2| = 12 - 4 \\[3ex] This\:\: means\:\: that \\[3ex] -2p + 2 = 8 \:\:OR\:\: -(-2p + 2) = 8 \\[3ex] -2p + 2 = 8 \\[3ex] -2p = 8 - 2 \\[3ex] -2p = 6 \\[3ex] p = -\dfrac{6}{2} \\[5ex] p = -3 \\[3ex] OR \\[3ex] -(-2p + 2) = 8 \\[3ex] Divide\:\: both\:\: sides\:\: by\:\: -1 \\[3ex] -2p + 2 = -8 \\[3ex] -2p = -8 - 2 \\[3ex] -2p = -10 \\[3ex] p = \dfrac{-10}{-2} \\[5ex] p = 5 \\[3ex] $ Check
Check for both values.
$ \underline{LHS} \\[3ex] |-2p + 2| + 4 \\[3ex] p = -3 \\[3ex] |-2(-3) + 2| + 4 \\[3ex] |6 + 2| + 4 \\[3ex] |8| + 4 \\[3ex] 8 + 4 \\[3ex] 12 \\[3ex] $ $p = -3$ is a solution



$ |-2p + 2| + 4 \\[3ex] p = 5 \\[3ex] |-2(5) + 2| + 4 \\[3ex] |-10 + 2| + 4 \\[3ex] |-8| + 4 \\[3ex] 8 + 4 \\[3ex] 12 \\[3ex] $ $p = 5$ is a solution 		$ \underline{RHS} \\[3ex] 12 $
(6.) $-2\left|3 - \dfrac{p}{3}\right| + 1 = -5$


$ -2\left|3 - \dfrac{p}{3}\right| + 1 = -5 \\[5ex] -2\left|3 - \dfrac{p}{3}\right| = -5 - 1 \\[5ex] -2\left|3 - \dfrac{p}{3}\right| = -6 \\[5ex] \left|3 - \dfrac{p}{3}\right| = \dfrac{-6}{-2} \\[5ex] \left|3 - \dfrac{p}{3}\right| = 3 \\[5ex] This\:\: means\:\: that \\[5ex] 3 - \dfrac{p}{3} = 3 \:\:OR\:\: -(3 - \dfrac{p}{3} = 3 \\[5ex] 3 - \dfrac{p}{3} = 3 \\[5ex] 3 - 3 = \dfrac{p}{3} \\[5ex] 0 = \dfrac{p}{3} \\[5ex] \dfrac{p}{3} = 0 \\[5ex] p = 0(3) \\[3ex] p = 0 \\[3ex] OR \\[3ex] -(3 - \dfrac{p}{3} = 3 \\[5ex] Divide\:\: both\:\: sides\:\: by\:\: -1 \\[3ex] 3 - \dfrac{p}{3} = -3 \\[5ex] 3 + 3 = \dfrac{p}{3} \\[5ex] 6 = \dfrac{p}{3} \\[5ex] \dfrac{p}{3} = 6 \\[5ex] p = 6(3) \\[3ex] p = 18 \\[3ex] $ Check
Check for both values.
$ \underline{LHS} \\[3ex] -2\left|3 - \dfrac{p}{3}\right| + 1 \\[5ex] p = 0 \\[3ex] -2\left|3 - \dfrac{0}{3}\right| + 1 \\[5ex] -2|3 - 0| + 1 \\[5ex] -2|3| + 1 \\[3ex] -2(3) + 1 \\[3ex] -6 + 1 \\[3ex] -5 \\[3ex] $ $p = 0$ is a solution



$ -2\left|3 - \dfrac{p}{3}\right| + 1 \\[5ex] p = 18 \\[3ex] -2\left|3 - \dfrac{18}{3}\right| + 1 \\[5ex] -2|3 - 6| + 1 \\[5ex] -2|-3| + 1 \\[3ex] -2(3) + 1 \\[3ex] -6 + 1 \\[3ex] -5 \\[3ex] $ $p = 18$ is a solution 		$ \underline{RHS} \\[3ex] -5 $
(7.)


(8.)