\(\Leftrightarrow\dfrac{20\left(5x+6\right)-35\left(3x+1\right)}{140}=\dfrac{28\left(x+16\right)}{140}\)
\(\Leftrightarrow20\left(5x+6\right)-35\left(3x+1\right)=28\left(x+16\right)\)
\(\Leftrightarrow100x+120-105x-35=28x+448\)
\(\Leftrightarrow-33x=363\)
\(\Leftrightarrow x=-11\)