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