помогите систему уравнений решить 2ху+у^2=28 и х+2у=10. б) х^2+y^2=50 и xy=7

Ответ
{ 2ху+у^2=28
{ х+2у=10 => x=10 - 2y
=>
2*(10-2y)*y + y^2 = 28
20 - 4y^2 + y^2 = 28
- 2y^2 = 8
y^2 = -4 - действительных корней нет, т. к. a^(2n) >= 0

{ х^2+y^2=50
{ xy=7 => x=7/y
=>
(7/y)^2 + y^2 = 50
49 + y^4 = 50y^2
y^4 - 50y^2 + 49 = 0
y^2 = t
t^2 - 50t + 49 = 0
t1 + t2 = 50
t1 * t2 = 49 =>
t1 = 1 => y^2=t1=1 => y1=+1, y2=-1
t2 = 49 => y^2=t2=49 => y3=+7, y4=-7
Итого получено 4 значения y => из условия:
xy = 7 => x=7/y =>
x1 = 7/y1 = 7/1 = 7
x2 = 7/y2 = 7/(-1) = -7
x3 = 7/y3 = 7/7 = 1
x4 = 7/y4 = 7/ (-7) = -1
Ответ:
при х1= 7 => y1= 1
при х2= -7 => y2= -1
при х3= 1 => y3= +7
при х4= -1 => y4= -7

Ответ. 1. 2*х*у+у^2=28; х+2у=10; y*(2*x+y)=28; 2*x+y=20-3*y; y*(20-3*y)=28; 3*y^2-20*y+28=0; y1=14/3; y2=2; x1=2/3; x2=6;
2. х^2+y^2=50; x*y=7; (x+y)^2=64; (x-y)^2=36; 1). x+y=8; y=8-x; (8-x)*x=7; x^2-8*x+7=0; x1=7; x2=1; y1=1; y2=7;
2). x+y=-8; y=-8-x; -x*(8+x)=7; x^2+8*x+7=0; x3=-7; x4=-1; y3=-1; y4=-7; 3). x-y=6; y=x-6; x*(x-6)=7; x^2-6*x-7=0; x5=7; x6=-1;y5=1; y6=-7;
4). x-y=-6; y=x+6; x*(x+6)=7; x^2+6*x-7=0; x7=-7; x8=1; y7=-1; y8=7;
(7;1); (1;7); (-7;-1); (-1;-7);