Odpowiedź:
[tex]\begin{cases}(x+2)^2-(x-1)^2=2y+3\\(y+1)^2-(y-3)^2=16x-8\end{cases}\\\\\\\begin{cases}x^2+4x+4-(x^2-2x+1)=2y+3\\y^2+2y+1-(y^2-6y+9)=16x-8\end{cases}\\\\\\\begin{cases}x^2+4x+4-x^2+2x-1=2y+3\\y^2+2y+1-y^2+6y-9=16x-8\end{cases}\\\\\\\begin{cases}6x+3=2y+3\\8y-8=16x-8\end{cases}\\\\\\\begin{cases}6x-2y=3-3\\-16x+8y=-8+8\end{cases}[/tex]
[tex]\begin{cases}6x-2y=0\\-16x+8y=0\ \ /:4\end{cases}\\\\\\+\begin{cases}6x-2y=0\\-4x+2y=0\end{cases}\\---------\\2x=0\ \ /:2\\x=0\\\\\\6*0-2y=0\\\\-2y=0\ \ /:(-2)\\\\y=0\\\\\begin{cases}x=0\\y=0\end{cases}[/tex]
