\(\left\{ \begin{array}{l}3{x^2}-2x = y,\\3x-2 = y\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}3{x^2}-2x = 3x-2,\\y = 3x-2\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}3{x^2}-5x + 2 = 0,\\y = 3x-2.\end{array} \right.\)
\(3{x^2}-5x + 2 = 0;\,\,\,\,\,\,\,\,D = {\left( {-5} \right)^2}-4 \cdot 3 \cdot 2 = 1;\,\,\,\,\,\,\,\,\,\left[ \begin{array}{l}x = \dfrac{{5 + 1}}{{2 \cdot 3}} = 1,\\x = \dfrac{{5-1}}{{2 \cdot 3}} = \dfrac{2}{3}.\end{array} \right.\)
Если \(x = 1,\) то \(y = 3 \cdot 1-2 = 1.\)
Если \(x = \dfrac{2}{3},\) то \(y = 3 \cdot \dfrac{2}{3}-2 = 0.\)
Ответ: \(\left( {1;\,1} \right),\,\,\;\left( {\dfrac{2}{3};\,0} \right).\)