1 способ
\(\left\{ \begin{array}{l}2{x^2} + y = 9,\\3{x^2}-y = 11\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}y = 9-2{x^2},\\y = 3{x^2}-11\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}3{x^2}-11 = 9-2{x^2},\\y = 3{x^2}-11\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}5{x^2} = 20,\\y = 3{x^2}-11\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}{x^2} = 4,\\y = 3{x^2}-11\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}\left[ {\begin{array}{*{20}{c}}{x = -2,}\\{x = 2,\,\,\,}\end{array}} \right.\\y = 3{x^2}-11\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ \begin{array}{l}\left\{ {\begin{array}{*{20}{c}}{x = -2,}\\{y = 1,\,\,}\end{array}} \right.\\\left\{ {\begin{array}{*{20}{c}}{x = 2,\,}\\{y = 1.}\end{array}} \right.\end{array} \right.\)
Ответ: \(\left( {-2;\,1} \right),\,\,\;\left( {2;\,1} \right).\)
2 способ
Прибавим к первому уравнению системы второе:
\(2{x^2} + y + 3{x^2}-y = 9 + 11\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,5{x^2} = 20\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,{x^2} = 4\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ \begin{array}{l}x = -2,\\x = 2.\end{array} \right.\)
Подставим найденные значения x в первое уравнение исходной системы.
Если \(x = 2,\) то \(2 \cdot {2^2} + y = 9\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,y = 1.\)
Если \(x = -2,\) то \(2 \cdot {\left( {-2} \right)^2} + y = 9\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,y = 1.\)
Ответ: \(\left( {-2;\,1} \right),\,\,\;\left( {2;\,1} \right).\)