Answer
$(-\infty,\dfrac{-7}{3} ) \cup (5, \infty)$
Work Step by Step
Here, $|3x-4| \gt 11$
As per definition of absolute value, we have $|x|=\pm x$
$3x-4 \gt 11$ and $3x-4 \gt -11$
Thus, $3x \gt 11+4$ and $3x \gt -11+4$
or, $x\gt 5$ and $ x \lt \dfrac{-7}{3}$
Hence, our solution set is: $(-\infty,\dfrac{-7}{3} ) \cup (5, \infty)$
