Giải các hệ phương trình sau bằng phương pháp cộng đại số

a) \(\) \(\left\{ \begin{array}{l}x\sqrt 2  - 3y = 1\\2x + y\sqrt 2  =  - 2\end{array} \right.\)

b) \(\left\{ \begin{array}{l}5x\sqrt 3  + y = 2\sqrt 2 \\x\sqrt 6  - y\sqrt 2  = 2\end{array} \right.\)

 e) \(\left\{ \begin{array}{l}0,3x + 0,5y = 3\\1,5x - 2y = 1,5\end{array} \right.\)

\(\begin{array}{l}\left\{ \begin{array}{l}1,2x + 2y = 12\\1,5x - 2y = 1,5\end{array} \right.\\\left\{ \begin{array}{l}1,2x + 2y = 12\\2,7x = 13,5\end{array} \right.\\\left\{ \begin{array}{l}x = 5\\y = 3\end{array} \right.\end{array}\)

Hệ có nghiệm duy nhất \(\left( {5\,,\,3} \right)\)

a) \[\left\{ \begin{array}{l} - 5{\rm{x}} + 2y = 4\\6{\rm{x}} - 3y =  - 7\end{array} \right.\]

\[\begin{array}{l}\left\{ \begin{array}{l} - 15x + 6y = 12\\12x - 6y =  - 14\end{array} \right.\\\left\{ \begin{array}{l} - 15x + 6y = 12\\ - 3x =  - 2\end{array} \right.\\\left\{ \begin{array}{l}x = \frac{2}{3}\\y = \frac{{11}}{3}\end{array} \right.\end{array}\]

Hệ có nghiệm duy nhất \[\left( {\frac{2}{3};\,\frac{{11}}{3}} \right)\]

b) \[\left\{ \begin{array}{l}2x - 3y = 11\\ - 4x + 6y = 5\end{array} \right.\]

 \[\begin{array}{l}\left\{ \begin{array}{l}4x - 6y = 22\\ - 4x + 6y = 5\end{array} \right.\\\left\{ \begin{array}{l}4x - 6y = 22\\0x + 0y = 27\end{array} \right.\end{array}\]

Hệ vô nghiệm

c) \[\left\{ \begin{array}{l}3x - 2y = 10\\x - \frac{2}{3}y = 3\frac{1}{3}\end{array} \right.\]

 \[\begin{array}{l}\left\{ \begin{array}{l}3x - 2y = 10\\x - \frac{2}{3}y = 3\frac{1}{3}\end{array} \right.\\\left\{ \begin{array}{l}3x - 2y = 10\\3x - 2y = 10\end{array} \right.\\3x - 2y = 10\\y = \frac{1}{2}\left( {3x - 10} \right)\end{array}\]

Hệ có vố số nghiệm: \[\left\{ \begin{array}{l}y = \frac{1}{2}\left( {3x - 10} \right)\\x \in \mathbb{R}\end{array} \right.\]