\(\left\{{}\begin{matrix}mx-y=1\\x+my=m+6\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}m^2x-my=m\\x+my=m+6\end{matrix}\right.\) \(\left(m\ne0\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(m^2+1\right)x=2m+6\\mx-y=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2m+6}{m^2+1}\\y=mx-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2m+6}{m^2+1}\\y=m.\dfrac{2m+6}{m^2+1}-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2m+6}{m^2+1}\\y=\dfrac{m^2+6m-1}{m^2+1}\end{matrix}\right.\)
Theo đề bài :
\(3x-y=1\)
\(\Leftrightarrow3.\dfrac{2m+6}{m^2+1}-\dfrac{m^2+6m-1}{m^2+1}=1\)
\(\Leftrightarrow\dfrac{19-m^2}{m^2+1}=1\)
\(\Leftrightarrow19-m^2=m^2+1\)
\(\Leftrightarrow2m^2=18\)
\(\Leftrightarrow m^2=9\)
\(\Leftrightarrow\left[{}\begin{matrix}m=3\\m=-3\end{matrix}\right.\) thỏa mãn yêu cầu của đề bài
