Odpowiedź:
[tex]a)\\\\\begin{cases}xy=(x+4)(y-2)\\(x-3)(y+3)=yx\end{cases}\\\\\\\begin{cases}xy=xy-2x+4y-8\\xy+3x-3y-9=xy\end{cases}\\\\\\\begin{cases}xy-xy+2x-4y=-8\\xy+3x-3y-xy=9\end{cases}\\\\\\\begin{cases}2x-4y=-8\ \ /:(-2)\\3x-3y=9\ \ /:3\end{cases}\\\\\\+\begin{cases}-x+2y=4\\x-y=3\end{cases}\\--------\\y=7\\\\\\x-7=3\\\\x=3+7\\\\x=10\\\\\begin{cases}x=10\\y=7\end{cases}[/tex]
[tex]b)\\\\\begin{cases}(y+4)(x-2)=xy\\(y-3)(x+3)=xy\end{cases}\\\\\\\begin{cases}xy-2y+4x-8=xy\\xy+3y-3x-9=xy\end{cases}\\\\\\\begin{cases}xy-2y+4x-xy=8\\xy+3y-3x-xy=9\end{cases}\\\\\\\begin{cases}-2y+4x=8\ \ /:2\\3y-3x=9\ \ /:3\end{cases}\\\\\\+\begin{cases}-y+2x=4\\y-x=3\end{cases}\\--------\\x=7\\\\\\y-7=3\\y=3+7\\\\y=10\\\\\begin{cases}x=7\\y=10\end{cases}[/tex]
