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# Solved Examples - Absolute Value Equations

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.
So, you should try to solve each question correctly and timely.
So, it is not just solving a question correctly, but solving it correctly on time.
Please ensure you attempt all ACT questions.
There is no "negative" penalty for any wrong answer.

For JAMB and CMAT Students
Calculators are not allowed. So, the questions are solved in a way that does not require a calculator.

Solve the absolute value equations as applicable.
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.) $26 - |8x - 3| = 6$

$26 - |8x - 3| = 6 \\[3ex] 26 - 6 = |8x - 3| \\[3ex] 20 = |8x - 3| \\[3ex] |8x - 3| = 20 \\[3ex] This\:\: means\:\: that \\[3ex] 8x - 3 = 20 \:\:OR\:\: -(8x - 3) = 20 \\[3ex] 8x - 3 = 20 \\[3ex] 8x = 20 + 3 \\[3ex] 8x = 23 \\[3ex] x = \dfrac{23}{8} \\[5ex] OR \\[3ex] -(8x - 3) = 20 \\[3ex] Divide\:\: both\:\: sides\:\: by\:\: -1 \\[3ex] 8x - 3 = -20 \\[3ex] 8x = -20 + 3 \\[3ex] 8x = -17 \\[3ex] x = -\dfrac{17}{8} \\[5ex]$ Check
Check for both values.
 $\underline{LHS} \\[3ex] 26 - |8x - 3| \\[3ex] x = \dfrac{23}{8} \\[5ex] 26 - \left|8\left(\dfrac{23}{8}\right) - 3\right| \\[5ex] 26 - |23 - 3| \\[3ex] 26 - |20| \\[3ex] 26 - 20 \\[3ex] 6 \\[3ex]$ $x = \dfrac{23}{8}$ 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 $\underline{RHS} \\[3ex] 6$
(3.) $|-2p + 2| + 4 = 12$

$|-2p + 2| + 4 = 12 \\[3ex] |-2p + 2| = 12 - 4 \\[3ex] This\:\: means\:\: that \\[3ex] -2p + 2 = 8 \:\:OR\:\: -(-2p + 2) = 8 \\[3ex] -2p + 2 = 8 \\[3ex] -2p = 8 - 2 \\[3ex] -2p = 6 \\[3ex] p = -\dfrac{6}{2} \\[5ex] p = -3 \\[3ex] OR \\[3ex] -(-2p + 2) = 8 \\[3ex] Divide\:\: both\:\: sides\:\: by\:\: -1 \\[3ex] -2p + 2 = -8 \\[3ex] -2p = -8 - 2 \\[3ex] -2p = -10 \\[3ex] p = \dfrac{-10}{-2} \\[5ex] p = 5 \\[3ex]$ 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 $\underline{RHS} \\[3ex] 12$
(4.) $-2\left|3 - \dfrac{p}{3}\right| + 1 = -5$

$-2\left|3 - \dfrac{p}{3}\right| + 1 = -5 \\[5ex] -2\left|3 - \dfrac{p}{3}\right| = -5 - 1 \\[5ex] -2\left|3 - \dfrac{p}{3}\right| = -6 \\[5ex] \left|3 - \dfrac{p}{3}\right| = \dfrac{-6}{-2} \\[5ex] \left|3 - \dfrac{p}{3}\right| = 3 \\[5ex] This\:\: means\:\: that \\[5ex] 3 - \dfrac{p}{3} = 3 \:\:OR\:\: -(3 - \dfrac{p}{3} = 3 \\[5ex] 3 - \dfrac{p}{3} = 3 \\[5ex] 3 - 3 = \dfrac{p}{3} \\[5ex] 0 = \dfrac{p}{3} \\[5ex] \dfrac{p}{3} = 0 \\[5ex] p = 0(3) \\[3ex] p = 0 \\[3ex] OR \\[3ex] -(3 - \dfrac{p}{3} = 3 \\[5ex] Divide\:\: both\:\: sides\:\: by\:\: -1 \\[3ex] 3 - \dfrac{p}{3} = -3 \\[5ex] 3 + 3 = \dfrac{p}{3} \\[5ex] 6 = \dfrac{p}{3} \\[5ex] \dfrac{p}{3} = 6 \\[5ex] p = 6(3) \\[3ex] p = 18 \\[3ex]$ 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 $\underline{RHS} \\[3ex] -5$