Question

Реши систему уравнений: {x+y=2

2y2+2xy+x2=68​

Answer 1

Ответ:

х=2-у

2у2+2у(2-у)+(2-у)2=68

2у2+4у-2у2+4-4у+у2

у2+4=68

у2=64

у=+-8

х1=2-(-8)=10

х2=2-8=-6

Answer 2

$\left \{ {{x+y=2} \atop {2y^{2}+2xy+x^{2} =68 }} \right. \\\\\left \{ {{x^{2}+2xy+2y^{2}=68 } \atop {(x+y)^{2} =2^{2} }} \right.\\\\-\left \{ {{x^{2}+2xy+2y^{2}=68 } \atop {x^{2}+2xy+y^{2}=4}} \right. \\---------\\y^{2}=64\\\\y_{1} =-8 \ ; \ y_{2} =8\\\\x_{1}=2-y_{1}=2-(-8)=2+8=10\\\\x_{2}=2-y_{2}=2-8=-6\\\\Otvet:\boxed{(10 \ ; \ -8) \ , \ (-6 \ ; \ 8)}$