Question

[Lizy]

\(\left\{{}\begin{matrix}3x+my=5\\x-2y=3\end{matrix}\right.\)

tìm m để hệ có nghiệm duy nhất (x;y) thỏa mãn 2x+y=1

Answer 1

[Nguyễn Việt Lâm]

Hệ có nghiệm duy nhất khi: \(\dfrac{3}{1}\ne\dfrac{m}{-2}\Rightarrow m\ne-6\)

Khi đó ta có:

\(\left\{{}\begin{matrix}3x+my=5\\x-2y=3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}6x+2my=10\\mx-2my=3m\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(m+6\right)x=3m+10\\y=\dfrac{x-3}{2}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{3m+10}{m+6}\\y=\dfrac{x-3}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3m+10}{m+6}\\y=\dfrac{-4}{m+6}\end{matrix}\right.\)

\(2x+y=1\Rightarrow\dfrac{2\left(3m+10\right)}{m+6}+\dfrac{-4}{m+6}=1\)

\(\Leftrightarrow\dfrac{6m+16}{m+6}=1\)

\(\Rightarrow6m+16=m+6\)

\(\Rightarrow m=-2\)