\(\left\{{}\begin{matrix}x+y+z=1\\y+z+t=2\\z+t+x=3\\t+x+y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1-y-z\\y+z+t=2\\z+t+1-y-z=3\\t+x+y=6\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=1-y-z\\y+z+t=2\\t-y=3\\t+1-y-z+y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1-y-z\\y+z+t=2\\t-y=3\\t-z=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=1-y-z\\t=2-y-z\\2-y-z-y=3\\2-y-z-z=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1-y-z\\t=2-y-z\\2y+z=-1\left(1\right)\\y+2z=-3\left(2\right)\end{matrix}\right.\)
Từ (1) và (2) ta có: \(\left\{{}\begin{matrix}2y+z=-1\\2y+4z=-6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2y+z=-1\\-3z=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\left(-1+\dfrac{5}{3}\right):2=\dfrac{1}{3}\\z=-\dfrac{5}{3}\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=1-\dfrac{1}{3}-\left(-\dfrac{5}{3}\right)=\dfrac{7}{3}\\t=2-\dfrac{1}{3}-\left(-\dfrac{5}{3}\right)=\dfrac{10}{3}\end{matrix}\right.\)
