Answer
$x\le-\dfrac{11}{3} \text{ or } x\ge\dfrac{19}{3}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To solve the given inequality, $
|3x-4|\ge15
,$ use the definition of a greater than (greater than or equal to) absolute value inequality. Then use the properties of inequality to isolate the variable in each resulting equation.
$\bf{\text{Solution Details:}}$
Since for any $c\gt0$, $|x|\gt c$ implies $x\gt c \text{ or } x\lt-c$ (which is equivalent to $|x|\ge c$ implies $x\ge c \text{ or } x\le-c$), the inequality above is equivalent to
\begin{array}{l}\require{cancel}
3x-4\ge15
\\\\\text{OR}\\\\
3x-4\le-15
.\end{array}
Using the properties of inequality to isolate the variable in each equation results to
\begin{array}{l}\require{cancel}
3x-4\ge15
\\\\
3x\ge15+4
\\\\
3x\ge19
\\\\
x\ge\dfrac{19}{3}
\\\\\text{OR}\\\\
3x-4\le-15
\\\\
3x\le-15+4
\\\\
3x\le-11
\\\\
x\le-\dfrac{11}{3}
.\end{array}
Hence, the solution set is $
x\le-\dfrac{11}{3} \text{ or } x\ge\dfrac{19}{3}
.$
