Задача 28. Решите систему уравнений: \(\left\{ \begin{array}{l}2{x^2} + 3{y^2} = 21,\\6{x^2} + 9{y^2} = 21x.\end{array} \right.\)
Ответ
ОТВЕТ: \(\left( {3;\,-1} \right),\,\,\,\;\left( {3;\,1} \right).\)
Решение
Разделим обе части второго уравнения на 3. Тогда система уравнений примет вид:
\(\left\{ \begin{array}{l}2{x^2} + 3{y^2} = 21,\\2{x^2} + 3{y^2} = 7x\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}7x = 21,\\2{x^2} + 3{y^2} = 21\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}x = 3,\\2{x^2} + 3{y^2} = 21\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}x = 3,\\2 \cdot {3^2} + 3{y^2} = 21\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}x = 3,\\{y^2} = 1\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}x = 3,\\\left[ \begin{array}{l}y = -1,\\y = 1\end{array} \right.\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ \begin{array}{l}\left\{ \begin{array}{l}x = 3,\\y = -1,\end{array} \right.\\\left\{ \begin{array}{l}x = 3,\\y = 1.\end{array} \right.\end{array} \right.\)
Ответ: \(\left( {3;\,-1} \right),\,\,\,\;\left( {3;\,1} \right).\)