\(\left\{ \begin{array}{l}2{x^2}-5x = y,\\2x-5 = y\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}2{x^2}-5x = 2x-5,\\y = 2x-5\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}2{x^2}-7x + 5 = 0,\\y = 2x-5.\end{array} \right.\)
\(2{x^2}-7x + 5 = 0;\,\,\,\,\,\,\,\,D = {\left( {-7} \right)^2}-4 \cdot 2 \cdot 5 = 9;\,\,\,\,\,\,\,\,\,\left[ \begin{array}{l}x = \dfrac{{7 + 3}}{{2 \cdot 2}} = \dfrac{5}{2},\\x = \dfrac{{7-3}}{{2 \cdot 2}} = 1.\end{array} \right.\)
Если \(x = \dfrac{5}{2},\) то \(y = 2 \cdot \dfrac{5}{2}-5 = 0.\)
Если \(x = 1,\) то \(y = 2 \cdot 1-5 = -3.\)
Ответ: \(\left( {1;\,-3} \right),\,\,\;\left( {\dfrac{5}{2};\,0} \right).\)