Системни уравнения от 2 степен

[Boco96]

Как се решават уравнения от този тип?Моля дайте решение с малко обяснения

|(x-3)(y-x-2)=0
|x^2 + y^2 - 3x + 4y=12

Благодаря предварително!

[ammornil]

[tex]\begin{array}{|l} (x-3).(y-x-2)=0 \\ x^2+y^2-3.x+4.y=12\end{array}[/tex]

[tex]\Rightarrow \begin{array}{|l} (x-3)=0 \\ x^2+y^2-3.x+4.y=12 \end{array}[/tex] [tex]\cup \begin{array}{|l}(y-x-2)=0 \\ x^2+y^2-3.x+4.y=12\end{array}[/tex]

[tex]\Rightarrow \begin{array}{|l} x=3 \\ 3^2+y^2-3.3+4.y=12 \end{array}[/tex] [tex]\cup \begin{array}{|l}y=x+2 \\ x^2+(x+2)^2-3.x+4.(x+2)=12\end{array}[/tex]

[tex]\Rightarrow \begin{array}{|l} x=3 \\ y^2+4.y-12=0 \end{array}[/tex] [tex]\cup \begin{array}{|l}y=x+2 \\ x^2+x^2+4.x+4-3.x+4.x+8-12=0\end{array}[/tex]

[tex]\Rightarrow \begin{array}{|l} x=3 \\ y_{1,2}=-2 \pm \sqrt{2^{2}+12} \end{array}[/tex] [tex]\cup \begin{array}{|l}y=x+2 \\ 2.x^2+5.x=0\end{array}[/tex]

[tex]\Rightarrow \begin{array}{|l} x=3 \\ y_{1,2}=-2 \pm 4 \end{array}[/tex] [tex]\cup \begin{array}{|l}y=x+2 \\ x.(2.x+5)=0\end{array}[/tex]

[tex]\Rightarrow \begin{array}{|l} x_{_{1}}=3 \\ y_1=-6 \end{array}[/tex] [tex]\cup \begin{array}{|l} x_2=3 \\ y_2=2 \end{array}[/tex] [tex]\cup \begin{array}{|l}y=x+2 \\ x_3=0\end{array}[/tex] [tex]\cup \begin{array}{|l} y=x+2 \\ x_4=-\frac{5}{2} \end{array}[/tex]

[tex]\Rightarrow \begin{array}{|l} x_1=3 \\ y_1=-6 \end{array}[/tex] [tex]\cup \begin{array}{|l} x_2=3 \\ y_2=2 \end{array}[/tex] [tex]\cup \begin{array}{|l} x_3=0 \\ y_3=2 \end{array}[/tex] [tex]\cup \begin{array}{|l} x_{4}=-\frac{5}{2} \\ y_4=-\frac{1}{2}\end{array}[/tex]