Решите систему уравнения: { X^2 + y^2 = 25 { xy - x - y = 5 покажите решение , пожалуйста

[latex] \left \{ {{ x^{2} +y^2=25} \atop {xy-x-y=5}} \right. [/latex] ; [latex] \left \{ {{ x^{2} +y^2-2xy+2xy=25} \atop {xy-x-y=5}} \right. [/latex] ; [latex] \left \{ {{ (x+y)^2-2xy=25} \atop {xy-(x+y)=5}} \right. [/latex] Введем замену пусть х+у=u, а ху=v [latex] \left \{ {{u^2-2v=25} \atop {v-u=5}} \right. [/latex] ; [latex] \left \{ {{u^2-2(5+u)=25} \atop {v=5+u}} \right. [/latex] ; [latex] \left \{ {{u^2-10-2u=25} \atop {v=5+u}} \right. [/latex] ; [latex]\left \{ {{u^2-2u-35=0} \atop {v=5+u}} \right.[/latex] ; [latex]D=2^2-4*1*35=4+140=144[/latex] ; [latex] u_{1}= \frac{2-12}{2}=-5 [/latex] ; [latex] u_{2}= \frac{2+12}{2}=7 [/latex] ; [latex] \left[\begin{array}{ccc} \left \{ {{ u_{1} =-5} \atop { v_{1} =5+u=5-5=0}} \right. \\ \left \{ {{ u_{2} =7} \atop { v_{2} =5+7=12}} \right. \end{array}\right[/latex] ; [latex] \left[\begin{array}{ccc} \left \{ {{ x+y =-5} \atop { xy =0}} \right. \\ \left \{ {{x+y=7} \atop { xy =12}} \right. \end{array}\right[/latex] ; [latex] \left[\begin{array}{ccc} \left \{ {{ x=-5-y} \atop { y(-5-y) =0}} \right. \\ \left \{ {{x=7-y} \atop { y(7-y) =12}} \right. \end{array}\right[/latex] ; [latex] \left[\begin{array}{ccc} \left \{ {{ x_{1} =-5-0=-5} \atop { y_{1}=0}} \right.\\ \left \{ {{ x_{2} =-5+5=0} \atop { y_{2} =-5}} \right. \\ \left \{ {{x=7-y} \atop { 7y-y^2 =12}} \right.\end{array}\right[/latex] Отдельно решим уравнение [latex]y^2-7y+12=0[/latex] [latex]D=7^2-4*1*12=49-48=1[/latex] [latex] y_{3}= \frac{7-1}{2}=3 [/latex] ; [latex] y_{4}= \frac{7+1}{2}=4 [/latex] [latex] \left[\begin{array}{ccc} \left \{ {{ x_{1} =-5} \atop { y_{1}=0}} \right.\\ \left \{ {{ x_{2} =0} \atop { y_{2} =-5}} \right. \\ \left \{ {{x_{3}=7-3}=4 \atop { y_{3}=3 }} \right.\\ \left \{ {{x_{4}=7-4=3} \atop {y_{4}=4}} \right. \end{array}\right[/latex] Ответ:{(-5;0), (0;-5), (4;3), (3;4)}