[tex]\displaystyle\bf\\\begin{cases}\sqrt{3}x-2\sqrt{2}y=-\sqrt{6}\\2\sqrt{3}x+\sqrt{2}y=3\sqrt{6}~~~\Big|~\cdot 2\end{cases}\\\\\\\begin{cases}\sqrt{3}x-2\sqrt{2}y=-\sqrt{6}\\4\sqrt{3}x+2\sqrt{2}y=6\sqrt{6}\end{cases}\\\\..........................................................~~~Adunam~ecuatiile.\\\\5\sqrt{3}x~~~~~~~/~~~=5\sqrt{6}\\\\x=\frac{5\sqrt{6}}{5\sqrt{3}}=\frac{\sqrt{6}}{\sqrt{3}}=\frac{\sqrt{3\cdot2}}{\sqrt{3}}=\frac{\sqrt{3}\cdot\sqrt{2}}{\sqrt{3}}=\sqrt{2}[/tex]
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[tex]\displaystyle\bf\\2\sqrt{3}x+\sqrt{2}y=3\sqrt{6}\\\\dar~~~x=\sqrt{2}\\\\ 2\sqrt{3}\sqrt{2}+\sqrt{2}y=3\sqrt{6}\\\\2\sqrt{3\cdot2}+\sqrt{2}y=3\sqrt{6}\\\\2\sqrt{6}+\sqrt{2}y=3\sqrt{6}\\\\\sqrt{2}y=3\sqrt{6}-2\sqrt{6}\\\\\sqrt{2}y=\sqrt{6}\\\\y=\frac{\sqrt{6}}{\sqrt{2}}=\frac{\sqrt{2\cdot3}}{\sqrt{2}}=\frac{\sqrt{2}\cdot\sqrt{3}}{\sqrt{2}}=\sqrt{3}\\\\\\Solutia:\\\\\\\boxed{\bf x=\sqrt{2} ;~~~y=\sqrt{3}}\\\\ Raspuns~corect~C)[/tex]
