Ta có: \(\dfrac{3x-2}{6}-5=\dfrac{3-2\left(x+7\right)}{4}\)
\(\Leftrightarrow\dfrac{3x-2}{6}-\dfrac{30}{6}=\dfrac{3-2x-14}{4}\)
\(\Leftrightarrow\dfrac{3x-32}{6}=\dfrac{-2x-11}{4}\)
\(\Leftrightarrow4\left(3x-32\right)=6\left(-2x-11\right)\)
\(\Leftrightarrow12x-128=-12x-66\)
\(\Leftrightarrow12x-128+12x+66=0\)
\(\Leftrightarrow24x-62=0\)
\(\Leftrightarrow24x=62\)
hay \(x=\dfrac{31}{12}\)
Vậy: \(S=\left\{\dfrac{31}{12}\right\}\)