Odpowiedź :
[tex]\begin{cases}x^2+y^2-4x-2y+4=0\\x-2y+2=0 \end{cases}\\\\\begin{cases}x^2+y^2-4x-2y+4=0\\x=2y-2 \end{cases}\\\\\begin{cases} (2y-2)^2+y^2-4 (2y-2)-2y+4=0\\x=2y-2 \end{cases} \\\\\begin{cases} 4y^2-8y+4+y^2-8y+8-2y+4 =0\\x=2y-2 \end{cases}\\\\ \begin{cases}5y^2-18y+16=0\\x=2y-2 \end{cases}[/tex]
[tex]5y^2-18y+16=0\\\\a=5,\ \ b=-18,\ \ c=16\\\\\Delta =b^2-4ac=(-18)^2-4*5*16= 324-320=4\\\\y_{1}=\frac{-b-\sqrt{\Delta }}{2a}=\frac{-(-18)-\sqrt{4}}{2*5}=\frac{18-2}{10}=\frac{16}{10}=\frac{8}{5}\\\\y_{2}=\frac{-b-\sqrt{\Delta }}{2a} =\frac{18+2}{10}=\frac{20}{10}=2[/tex]
[tex]\begin{cases}y=\frac{8}{5}\\x=2y-2\end{cases}\ \ \ lub\ \ \begin{cases}y=2\\x=2y-2\end{cases}\\\\\begin{cases}y=\frac{8}{5}\\x=2*\frac{8}{5}-2\end{cases}\ \ \ lub\ \ \begin{cases}y=2\\x=2*2-2\end{cases}\\\\\begin{cases}y=\frac{8}{5}\\x= \frac{16}{5}-2\end{cases}\ \ \ lub\ \ \begin{cases}y=2\\x=4-2\end{cases}\\\\\begin{cases}y=\frac{8}{5}\\x= \frac{16}{5}- \frac{10}{5}\end{cases}\ \ \ lub\ \ \begin{cases}y=2\\x= 2\end{cases} \\\\\begin{cases}y=\frac{8}{5}\\x= \frac{ 6}{5} \end{cases}\ \ \ lub\ \ \begin{cases}y=2\\x= 2\end{cases}[/tex]