\(\left(3x-1\right)\left(4x+3\right)=2\left(3x-1\right)\\ \Leftrightarrow12x^2+9x-4x-3=6x-2\\ \Leftrightarrow12x^2+9x-4x-6x=-2+3\\ \Leftrightarrow12x^2-x=1\\ \Leftrightarrow x\left(12x-1\right)=1\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\12x-1=1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{6}\end{matrix}\right.\)
\(\left(3x-1\right)\cdot\left(4x+3\right)=2\cdot\left(3x-1\right)\)
\(\Leftrightarrow\)\(\left(3x-1\right)\cdot\left(4x+3\right)-2\cdot\left(3x-1\right)=0\)
\(\Leftrightarrow\)\(\left(3x-1\right)\cdot\left(4x+3-2\right)=0\)
\(\Leftrightarrow\)\(\left(3x-1\right)\cdot\left(4x+1\right)=0\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}3x-1=0\\4x+1=0\end{matrix}\right.\)
\(\Leftrightarrow3x=0+1\)
\(\Leftrightarrow3x=1\)
\(\Leftrightarrow x=1\div3\)
\(\Leftrightarrow x=\dfrac{1}{3}\)
\(\Leftrightarrow4x+1=0\)
\(\Leftrightarrow4x=0-1\)
\(\Leftrightarrow4x=-1\)
\(\Leftrightarrow x=-1\div4\)
\(\Leftrightarrow x=-\dfrac{1}{4}\)
\(\Rightarrow S=\left[{}\begin{matrix}\dfrac{1}{3}\\-\dfrac{1}{4}\end{matrix}\right.\)
