Вопрос по решению 4-ой и 5-ой системы

Question 1

Question 2

4-ое уже решил

Question 3

Решите задачу:

$4)\; \left \{ {{xy+x+y=15} \atop {x^2y+y^x=54}} \right. \; \; \; \left \{ {{xy+x+y=15} \atop {xy(x+y)=54}} \right.\\\\Zamena;\;\; \; u=xy\; ,\; v=x+y\\\\ \left \{ {{u+v=15} \atop {u\cdot v=54}} \right. \; \; \to \; \; teor.\; Vieta\; :\; u=6\; ,\; v=9\; \; ili\; \; u=9,\; v=6\\\\ \left \{ {{x\cdot y=6} \atop {x+y=9}} \right. \; \left \{ {{x(9-x)=6} \atop {y=9-x}} \right. \; \left \{ {{x^2-9x+6=0} \atop {y=9-x}} \right. \\\\D=9^2-4\cdot 6=57\; ;\; x_{1,2}=\frac{9\pm \sqrt{57}}{2}\; ;$

$y_{1,2}=9-\frac{9\pm \sqrt{57}}{2}=\frac{9\mp \sqrt{57}}{2}\\\\Otvet:\; (\frac{9+\sqrt{57}}{2};\frac{9-\sqrt{57}}{2})\; ;\; (\frac{9-\sqrt{57}}{2};\frac{9+\sqrt{57}}{2})$

$5)\; \left \{ {{x+y+xy=7} \atop {x^2+y^2+xy=13}} \right. \; \; Zamena;\; u=x+y\; ,\; v=xy\; ,\\\\u^2=x^2+y^2+2xy=x^2+y^2+2v\; \to \; x^2+y^2=u^2-2v\\\\ \left \{ {{u+v=7} \atop {u^2-v=13}} \right. \; \oplus \left \{ {{u+v=7} \atop {u^2+u-20=0}} \right. \; \left \{ {{v=7-u} \atop {u_1=4,u_2=-5}} \right. \\\\ \left \{ {{v_1=3} \atop {u_1=4}} \right.\; \; ili\; \; \left \{ {{v_2=12} \atop {u_2=-5}} \right.$

$\left \{ {{x+y=4} \atop {xy=3}} \right. \; \left \{ {{y=4-x} \atop {x^2-4x+3=0}} \right. \; \to \; \; x_1=1,\; x_2=3\\\\y_1=4-1=3\; ,\; y_2=4-3=1\\\\ \left \{ {{x+y=-5} \atop {xy=12}} \right. \; \left \{ {{y=-x-5} \atop {x^2+5x+12=0}} \right. \; \left \{ {{y=-x-5} \atop {D=25-4\cdot 12\ \textless \ 0\; \to \; net\; reshenij}} \right. \\\\Otvet:\; (1,3)\; ;\; (3,1)\; .\\\\oo$