Question

X^2+x+xy=8
y^2+y+xy=4 решить систему уравнений

Answer 1

$left { {{x^2+x+xy=8} atop {y^2+y+xy=4}} right. oplus left { {{x^2+y^2+x+y+2xy=12} atop {x^2+x+xy=8}} right. ; left { {{(x+y)^2+(x+y)-12=0} atop {x^2+x+xy=8}} right. \\ t=x+y; ,; ; t^2+t-12=0; ,; ; t_1=-4; ,; t_2=3; ; (teorema; Vieta)\\a); ; left { {{x+y=-4} atop {x^2+x+xy=8}} right. ; left { {{y=-x-4} atop {x^2+x+x(-x-4)=8}} right. ; left { {{y=-x-4} atop {x^2+x-x^2-4x=8}} right. ; left { {{y=-x-4} atop {-3x=8}} right.$

$left { {{y=-frac{4}{3}} atop {x=-frac{8}{3}}} right. \\b); ; left { {{x+y=3} atop {x^2+x+xy=8}} right. ; left { {{y=3-x} atop {x^2+x+x(3-x)=8}} right. ; left { {{y=3-x} atop {x^2+x+3x-x^2=8}} right. ; left { {{y=3-x} atop {4x=8}} right. \\left { {{y=1} atop {x=2}} right. \\Otvet:; ; (-frac{8}{3}, ,, -frac{4}{3}); ,; ; (2,1); .$