\(\left\{ \begin{array}{l}3{x^2}-4x = y,\\3x-4 = y\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}3{x^2}-4x = 3x-4,\\y = 3x-4\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}3{x^2}-7x + 4 = 0,\\y = 3x-4.\end{array} \right.\)
\(3{x^2}-7x + 4 = 0;\,\,\,\,\,\,\,\,D = {\left( {-7} \right)^2}-4 \cdot 3 \cdot 4 = 1;\,\,\,\,\,\,\,\,\,\left[ \begin{array}{l}x = \dfrac{{7 + 1}}{{2 \cdot 3}} = \dfrac{4}{3},\\x = \dfrac{{7-1}}{{2 \cdot 3}} = 1.\end{array} \right.\)
Если \(x = \dfrac{4}{3},\) то \(y = 3 \cdot \dfrac{4}{3}-4 = 0.\)
Если \(x = 1,\) то \(y = 3 \cdot 1-4 = -1.\)
Ответ: \(\left( {1;\,-1} \right),\,\,\;\left( {\dfrac{4}{3};\,0} \right).\)