1 способ
\(\left\{ \begin{array}{l}{x^2} + y = 5,\\6{x^2}-y = 2\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}y = 5-{x^2},\\y = 6{x^2}-2\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}6{x^2}-2 = 5-{x^2},\\y = 6{x^2}-2\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}7{x^2} = 7,\\y = 6{x^2}-2\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}{x^2} = 1,\\y = 6{x^2}-2\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}\left[ {\begin{array}{*{20}{c}}{x = -1,}\\{x = 1,\,\,\,}\end{array}} \right.\\y = 6{x^2}-2\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ \begin{array}{l}\left\{ {\begin{array}{*{20}{c}}{x = -1,}\\{y = 4,\,\,}\end{array}} \right.\\\left\{ {\begin{array}{*{20}{c}}{x = 1,\,}\\{y = 4.}\end{array}} \right.\end{array} \right.\)
Ответ: \(\left( {-1;\,4} \right),\,\,\;\left( {1;\,4} \right).\)
2 способ
Прибавим к первому уравнению системы второе:
\({x^2} + y + 6{x^2}-y = 5 + 2\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,7{x^2} = 7\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,{x^2} = 1\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ \begin{array}{l}x = -1,\\x = 1.\end{array} \right.\)
Подставим найденные значения x в первое уравнение исходной системы.
Если \(x = 1,\) то \({1^2} + y = 5\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,y = 4.\)
Если \(x = -1,\) то \({\left( {-1} \right)^2} + y = 5\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,y = 4.\)
Ответ: \(\left( {-1;\,4} \right),\,\,\;\left( {1;\,4} \right).\)