Odpowiedź:
[tex]a)\\\\\begin{cases}y=(2x-1)^2-(2x+3)^2\\(2y-1)(2y+1)+96=4y(y+3)+x\end{cases}\\\\\\\begin{cases}y=4x^2-4x+1-(4x^2+12x+9)\\4y^2-1+96=4y^2+12y+x\end{cases}\\\\\\\begin{cases}y=4x^2-4x+1-4x^2-12x-9\\4y^2+95=4y^2+12y+x\end{cases}\\\\\\\begin{cases}y=-16x-8\\4y^2-4y^2-12y-x=-95\end{cases}[/tex]
[tex]\begin{cases}16x+y=-8\ \ /\cdot12\\-x-12y=-95\end{cases}\\\\\\+\begin{cases}192x+12y=-96\\-x-12y=-95\end{cases}\\-----------\\191x=-191\ \ /:191\\x=-1\\\\\\16\cdot(-1)+y=-8\\\\-16+y=-8\\\\y=-8+16\\\\y=8\\\\\begin{cases}x=-1\\y=8\end{cases}[/tex]
[tex]b)\\\\\begin{cases}0,3x-0,7y=5\ \ /\cdot10\\4x+7y=-15\end{cases}\\\\\\+\begin{cases}3x-7y=50\\4x+7y=-15\end{cases}\\---------\\7x=35\ \ /:7\\x=5\\\\\\4\cdot5+7y=-15\\\\20+7y=-15\\\\7y=-15-20\\\\7y=-35\ \ /:7\\\\y=-5\\\\\begin{cases}x=5\\y=-5\end{cases}[/tex]
