Đặt \(\left\{{}\begin{matrix}\sqrt{3x+y}=a\ge0\\\sqrt{2x+y}=b\ge0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}3x+y=a^2\\2x+y=b^2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=a^2-b^2\\y=3b^2-2a^2\end{matrix}\right.\)
Hệ đã cho trở thành:
\(\left\{{}\begin{matrix}a+b=3\\b-3a^2+4b^2=15\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=3-b\\b-3\left(3-b\right)^2+4b^2-15=0\end{matrix}\right.\)
\(\Rightarrow b^2+19b-42=0\Rightarrow\left[{}\begin{matrix}b=2\Rightarrow a=1\\b=-21< 0\left(l\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=a^2-b^2=-3\\y=3b^2-2a^2=10\end{matrix}\right.\)
