Question

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Answer 1

Ta có: \(\left\{{}\begin{matrix}\left(x+2\right)\left(y-3\right)=xy\\\left(x-1\right)\left(y-2\right)=xy\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}xy-3x+2y-6=xy\\xy-2x-y+2=xy\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}-3x+2y=6\\-2x-y=-2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}-3x+2y=6\\-4x-2y=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-3x+2y-4x-2y=6-4\\2x+y=2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}-7x=2\\y=2-2x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{2}{7}\\y=2-2\cdot\dfrac{-2}{7}=2+\dfrac{4}{7}=\dfrac{18}{7}\end{matrix}\right.\)