\(\left\{ \begin{array}{l}5{x^2}-9x = y,\\5x-9 = y\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}5{x^2}-9x = 5x-9,\\y = 5x-9\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}5{x^2}-14x + 9 = 0,\\y = 5x-9.\end{array} \right.\)
\(5{x^2}-14x + 9 = 0;\,\,\,\,\,\,\,\,D = {\left( {-14} \right)^2}-4 \cdot 5 \cdot 9 = 16;\,\,\,\,\,\,\,\,\,\left[ \begin{array}{l}x = \dfrac{{14 + 4}}{{2 \cdot 5}} = \dfrac{9}{5},\\x = \dfrac{{14-4}}{{2 \cdot 5}} = 1.\end{array} \right.\)
Если \(x = \dfrac{9}{5},\) то \(y = 5 \cdot \dfrac{9}{5}-9 = 0.\)
Если \(x = 1,\) то \(y = 5 \cdot 1-9 = -4.\)
Ответ: \(\left( {1;\,-4} \right),\,\,\;\left( {\dfrac{9}{5};\,0} \right).\)