Answer
$x=\left\{ \dfrac{31}{3},\dfrac{41}{3} \right\}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To solve the given equation, $
-3|x-12|=-5
,$ isolate first the absolute value expression. Then use the definition of absolute value equality. Do checking of the solution/s.
$\bf{\text{Solution Details:}}$
Using the properties of equality, the given equation is equivalent to
\begin{array}{l}\require{cancel}
-3|x-12|=-5
\\\\
|x-12|=\dfrac{-5}{-3}
\\\\
|x-12|=\dfrac{5}{3}
.\end{array}
Since for any $c\gt0$, $|x|=c$ implies $x=c \text{ or } x=-c,$ the equation above is equivalent to
\begin{array}{l}\require{cancel}
x-12=\dfrac{5}{3}
\\\\\text{OR}\\\\
x-12=-\dfrac{5}{3}
.\end{array}
Solving each equation results to
\begin{array}{l}\require{cancel}
x-12=\dfrac{5}{3}
\\\\
x=\dfrac{5}{3}+12
\\\\
x=\dfrac{5}{3}+\dfrac{36}{3}
\\\\
x=\dfrac{41}{3}
\\\\\text{OR}\\\\
x-12=-\dfrac{5}{3}
\\\\
x=-\dfrac{5}{3}+12
\\\\
x=-\dfrac{5}{3}+\dfrac{36}{3}
\\\\
x=\dfrac{31}{3}
.\end{array}
If $x=\dfrac{41}{3},$ then
\begin{array}{l}\require{cancel}
-3|x-12|=-5?
\\\\
-3\left| \dfrac{41}{3}-12 \right|=-5?
\\\\
-3\left| \dfrac{41}{3}-\dfrac{36}{3} \right|=-5?
\\\\
-3\left| \dfrac{5}{3} \right|=-5?
\\\\
-3\left( \dfrac{5}{3} \right)=-5?
\\\\
-5=-5
\text{ (TRUE)}
.\end{array}
If $x=\dfrac{31}{3},$ then
\begin{array}{l}\require{cancel}
-3|x-12|=-5?
\\\\
-3\left| \dfrac{31}{3}-12 \right|=-5?
\\\\
-3\left| \dfrac{31}{3}-\dfrac{36}{3} \right|=-5?
\\\\
-3\left| -\dfrac{5}{3} \right|=-5?
\\\\
-3\left( \dfrac{5}{3} \right)=-5?
\\\\
-5=-5
\text{ (TRUE)}
.\end{array}
Hence, $
x=\left\{ \dfrac{31}{3},\dfrac{41}{3} \right\}
.$
