\(\dfrac{x-1}{3}=\dfrac{2y-1}{4}=\dfrac{z+2}{5}=\dfrac{y+t+3}{6}=\dfrac{x-1+2y-1+z+2-y-t-3}{3+4+5-6}=\dfrac{\left(x+y+z-t\right)-3}{6}=\dfrac{1-3}{6}=-\dfrac{2}{3}\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-1=-2\\2y-1=-\dfrac{8}{3}\\z+2=-\dfrac{10}{3}\\y+t+3=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=-\dfrac{5}{6}\\z=-\dfrac{16}{3}\\t=-\dfrac{37}{6}\end{matrix}\right.\)
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