1 способ
\(\left\{ \begin{array}{l}5{x^2} + y = 12,\\9{x^2}-y = 2\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}y = 12-5{x^2},\\y = 9{x^2}-2\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}9{x^2}-2 = 12-5{x^2},\\y = 9{x^2}-2\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}14{x^2} = 14,\\y = 9{x^2}-2\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}{x^2} = 1,\\y = 9{x^2}-2\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}\left[ {\begin{array}{*{20}{c}}{x = -1,}\\{x = 1,\,\,\,}\end{array}} \right.\\y = 9{x^2}-2\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ \begin{array}{l}\left\{ {\begin{array}{*{20}{c}}{x = -1,}\\{y = 7,\,\,}\end{array}} \right.\\\left\{ {\begin{array}{*{20}{c}}{x = 1,\,}\\{y = 7.}\end{array}} \right.\end{array} \right.\)
Ответ: \(\left( {-1;\,7} \right),\,\,\;\left( {1;\,7} \right).\)
2 способ
Прибавим к первому уравнению системы второе:
\(5{x^2} + y + 9{x^2}-y = 12 + 2\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,14{x^2} = 14\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,{x^2} = 1\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ \begin{array}{l}x = -1,\\x = 1.\end{array} \right.\)
Подставим найденные значения x в первое уравнение исходной системы.
Если \(x = 1,\) то \(5 \cdot {1^2} + y = 12\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,y = 7.\)
Если \(x = -1,\) то \(5 \cdot {\left( {-1} \right)^2} + y = 12\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,y = 7.\)
Ответ: \(\left( {-1;\,7} \right),\,\,\;\left( {1;\,7} \right).\)