\(\left\{ \begin{array}{l}7{x^2}-5x = y,\\7x-5 = y\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}7{x^2}-5x = 7x-5,\\y = 7x-5\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}7{x^2}-12x + 5 = 0,\\y = 7x-5.\end{array} \right.\)
\(7{x^2}-12x + 5 = 0;\,\,\,\,\,\,\,\,D = {\left( {-12} \right)^2}-4 \cdot 7 \cdot 5 = 4;\,\,\,\,\,\,\,\,\,\left[ \begin{array}{l}x = \dfrac{{12 + 2}}{{2 \cdot 7}} = 1,\\x = \dfrac{{12-2}}{{2 \cdot 7}} = \dfrac{5}{7}.\end{array} \right.\)
Если \(x = 1,\) то \(y = 7 \cdot 1-5 = 2.\)
Если \(x = \dfrac{5}{7},\) то \(y = 7 \cdot \dfrac{5}{7}-5 = 0.\)
Ответ: \(\left( {1;\,2} \right),\,\,\;\left( {\dfrac{5}{7};\,0} \right).\)