1 способ
\(\left\{ \begin{array}{l}6{x^2} + y = 14,\\12{x^2}-y = 4\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}y = 14-6{x^2},\\y = 12{x^2}-4\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}12{x^2}-4 = 14-6{x^2},\\y = 12{x^2}-4\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}18{x^2} = 18,\\y = 12{x^2}-4\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}{x^2} = 1,\\y = 12{x^2}-4\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}\left[ {\begin{array}{*{20}{c}}{x = -1,}\\{x = 1,\,\,\,}\end{array}} \right.\\y = 12{x^2}-4\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ \begin{array}{l}\left\{ {\begin{array}{*{20}{c}}{x = -1,}\\{y = 8,\,\,}\end{array}} \right.\\\left\{ {\begin{array}{*{20}{c}}{x = 1,\,}\\{y = 8.}\end{array}} \right.\end{array} \right.\)
Ответ: \(\left( {-1;\,8} \right),\,\,\;\left( {1;\,8} \right).\)
2 способ
Прибавим к первому уравнению системы второе:
\(6{x^2} + y + 12{x^2}-y = 14 + 4\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,18{x^2} = 18\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,{x^2} = 1\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ \begin{array}{l}x = -1,\\x = 1.\end{array} \right.\)
Подставим найденные значения x в первое уравнение исходной системы.
Если \(x = 1,\) то \(6 \cdot {1^2} + y = 14\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,y = 8.\)
Если \(x = -1,\) то \(6 \cdot {\left( {-1} \right)^2} + y = 14\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,y = 8.\)
Ответ: \(\left( {-1;\,8} \right),\,\,\;\left( {1;\,8} \right).\)