Решите систему уравнений:
2x+y=-1
x^2+2y=3
————
y^2+xy=2
3y-x+7=0

{2x+y=-1|×(-2){x²+2y=3{-4х-2у=2{х²+2у=3-4х-2у+х²+2у=2+3х²-4х=5х²-4х-5=0По теореме Виета:х1+х2=-(-4)=4х1×х2=-5х1=-1х2=52х+у=-12×х1+у1=-1у1=-1-2×х1у1=-1-2×(-1)у1=-1+2у1=12×х2+у2=-12×5+у2=-110+у2=-1у2=-1-10у2=-11Ответ: (-1;1); (5;-11){y²+xy=2{3y-x+7=0х=(3у+7)у²+у×(3у+7)=2у²+3у²+7у-2=04у²+7у-2=0D=(-7)²-4×4×(-2)=49+32=81y1=(-7-√81)/2×4=(-7-9)/8=-16/8=-2y2=(-7+√81)/2×4=(-7+9)/8=(2/8)=1/4=0,25y1=-2y2=0,25x1=(3×y1+7)x1=3×(-2)+7x1=-6+7x1=1x2=(3×y2+7)x2=3×0,25+7x2=0,75+7x2=7,75Ответ: (1;-2); (7,75;0,25).