Odpowiedź:
[tex]\begin{cases}0,5y=3+2x\\x-2y=2\end{cases}\\\\\\\begin{cases}-2x+0,5y=3\ \ /\cdot4\\x-2y=2\end{cases}\\\\\\+\begin{cases}-8x+2y=12\\x-2y=2\end{cases}\\----------\\-7x=14\ \ /:(-7)\\x=-2\\\\\\-2-2y=2\\\\-2y=2+2\\\\-2y=4\ \ /:(-2)\\\\y=-2\\\\\\\begin{cases}x=-2\\y=-2\end{cases}[/tex]
