a) Ta có: \(\left\{{}\begin{matrix}2\left(x+y\right)=3x-y+7\\3\left(x-2y\right)=x+y+8\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+2y-3x+y=7\\3x-6y-x-y=8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-x+3y=7\\2x-7y=8\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-2x+6y=14\\2x-7y=8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-y=6\\-x+3y=7\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=-6\\x=3y-7=-18-7=-25\end{matrix}\right.\)
b) Ta có: \(\left\{{}\begin{matrix}-x+2y=-4\left(x-1\right)\\5x+2y=-\left(x+y\right)+8\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-x+2y+4x=4\\5x+2y+x+y=8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x+2y=4\\6x+3y=8\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}6x+4y=8\\6x+3y=8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=0\\3x=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{4}{3}\\y=0\end{matrix}\right.\)
