Ta có: \(3x-y=1\)\(\Leftrightarrow y=3x-1\) (1)
Thay (1) vào hệ phương trình, ta có:
\(\left\{{}\begin{matrix}mx-3x+1=1\\x+m\left(3x-1\right)=m+6\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}mx-3x=0\\x+3mx-2m=6\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}m=3\\x+3mx-2m=6\end{matrix}\right.\) \(\left(x\ne0\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}m=3\\x-3.3x-2.3=6\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}m=3\\-9x=12\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}m=3\\x=\frac{-3}{2}\\y=-\frac{11}{2}\end{matrix}\right.\)
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