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Solve the system: 3x-y=-1 2x-y+z=-6 x+4y-z=9
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I'd find a substitution for \(y\) or \(x\) from the first equation, then plug it into the second/third equations. Should be easy to solve for the remaining variables.
\[\begin{cases}3x-y=-1\\2x-y+z=-6\\x+4y-z=9\end{cases}~\Rightarrow~\begin{cases}\color{red}{y=3x+1}\\2x-\color{red}{(3x+1)}+z=-6\\x+4\color{red}{(3x+1)}-z=9\end{cases}\]
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