Câu 4:
a) Ta có: \(\left\{{}\begin{matrix}2x+y=1\\3x+4y=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}6x+3y=3\\6x+8y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-5y=5\\2x+y=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=-1\\2x-1=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=2\\y=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)
Vậy: hệ phương trình có nghiệm duy nhất là (x,y)=(1;-1)
a) \(\left\{{}\begin{matrix}2x+y=1\\3x+4y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=1-2x\\3x+4\left(1-2x\right)=-1\end{matrix}\right.\Leftrightarrow}\left\{{}\begin{matrix}y=1-2x\\-5x=-5\end{matrix}\right.\Leftrightarrow}\left\{{}\begin{matrix}y=-1\\x=1\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(1;-1\right)\)