Odpowiedź:
[tex]\begin{cases}3(x-1)(x+3)-2(y+1)(y+2)+2y^2=3(x^2-y)\\\frac{y-1}{4}=\frac{x}{2}-1\ \ /\cdot4\end{cases}\\\\\\\begin{cases}3(x^2+3x-x-3)-2(y^2+2y+y+2)+2y^2=3x^2-3y\\y-1=2x-4\end{cases}\\\\\\\begin{cases}3(x^2+2x-3)-2(y^2+3y+2)+2y^2=3x^2-3y\\y-2x=-4+1\end{cases}\\\\\\\begin{cases}3x^2+6x-9-2y^2-6y-4+2y^2=3x^2-3y\\-2x+y=-3\end{cases}\\\\\\\begin{cases}3x^2+6x-6y-3x^2+3y=9+4\\-2x+y=-3\end{cases}[/tex]
[tex]\begin{cases}6x-3y=13\\-2x+y=-3\ \ /\cdot3\end{cases}\\\\\\+\begin{cases}6x-3y=13\\-6x+3y=-9\end{cases}\\----------\\0=4\\\\Uklad\ \ r\'owna\'n\ \ sprzeczny\ \ nie\ \ ma\ \ rozwiazania[/tex]
