\(\left\{ \begin{array}{l}4{x^2}-5x = y,\\8x-10 = y\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}4{x^2}-5x = 8x-10,\\y = 8x-10\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}4{x^2}-13x + 10 = 0,\\y = 8x-10.\end{array} \right.\)
\(4{x^2}-13x + 10 = 0;\,\,\,\,\,\,\,\,D = {\left( {-13} \right)^2}-4 \cdot 4 \cdot 10 = 9;\,\,\,\,\,\,\,\,\,\left[ \begin{array}{l}x = \dfrac{{13 + 3}}{{2 \cdot 4}} = 2,\\x = \dfrac{{13-3}}{{2 \cdot 4}} = \dfrac{5}{4}.\end{array} \right.\)
Если \(x = 2,\) то \(y = 8 \cdot 2-10 = 6.\)
Если \(x = \dfrac{5}{4},\) то \(y = 8 \cdot \dfrac{5}{4}-10 = 0.\)
Ответ: \(\left( {2;\,6} \right),\,\,\;\left( {\dfrac{5}{4};\,0} \right).\)