Ta có : \(\left\{{}\begin{matrix}2x+y=3\\3x+\left(m-2\right)y=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3-y}{2}\\\dfrac{3\left(3-y\right)}{2}+\left(m-2\right)y=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3-y}{2}\\3\left(3-y\right)+2y\left(m-2\right)=8\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3-y}{2}\\y=\dfrac{1}{7-2m}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3-\dfrac{1}{7-2m}}{2}=\dfrac{10-3m}{7-2m}\\y=\dfrac{1}{2m-7}\end{matrix}\right.\)
Theo đề bài thì \(x+y=1\)
\(\Rightarrow\dfrac{10-3m}{7-2m}+\dfrac{1}{7-2m}=1\)
\(\Leftrightarrow10-3m+1=7-2m\)
\(\Leftrightarrow m=4\)
Vậy : \(m=4\)
