Question

Объясните пожалуйста как это решать дословно чтобы было понятно пж
Система
x2+y2=5,
xy=-2;
​

Answer 1

$\left \{ {{x^{2}+y^{2}=5} \atop {xy=-2}|*2} \right.\\\\+\left \{ {{x^{2}+y^{2}=5} \atop {2xy=-4}} \right.\\\\\left \{ {{x^{2}+2xy+y^{2}=1} \atop {xy=-2}} \right.\\\\\ \left \{ {{(x+y)^{2} =1} \atop {xy=-2}} \right.$

$1)\left \{ {{x+y=1} \atop {xy=-2}} \right.\\\\\left \{ {{x=1-y} \atop {(1-y)y=-2}} \right.\\\\\left \{ {{x=1-y} \atop {y^{2}-y-2=0 }} \right.\\\\y^{2}-y-2=0\\\\D=(-1)^{2}-4*(-2)=1+8=9=3^{2}\\\\y_{1}=\frac{1+3}{2}=2\\\\y_{2}=\frac{1-3}{2} =-1\\\\x_{1}=1-2=-1\\\\x_{2}=1-(-1)=2$

$2)\left \{ {{x+y=-1} \atop {xy=-2}} \right.\\\\\left \{ {{x=-y-1} \atop {(-y-1)y=-2}} \right.\\\\\left \{ {{x=-y-1} \atop {y^{2}+y-2=0 }} \right.\\\\y^{2}+y-2=0\\\\D=1^{2}-4*(-2)=1+8=9=3^{2}\\\\y_{3}=\frac{-1+3}{2}=1\\\\y_{4}=\frac{-1-3}{2}=-2\\\\x_{3}=-1-1=-2\\\\x_{4}=2-1=1$

Ответ : (- 1 ; - 2) , (2 ; - 1) , (- 2 ; - 1) , (  1 ; - 2)