Question

Решите систему уравнений x^2-3xy+y^2=-1 и 2x^2+5xy-y^2=17

Answer 1

$left { {{x^2-3xy+y^2=-1, |cdot 17} atop {2x^2+5xy-y^2=17}} right. ; oplus ; left { {{x^2-3xy+y^2=-1} atop {19x^2-46xy+16y^2=0}} right. \\19x^2-46xy+16y^2=0; |:y^2ne 0\\19cdot ( frac{x}{y} )^2-46cdot frac{x}{y} +16=0\\t= frac{x}{y} ; ,; ; 19t^2-46t+16=0; ,\\D/4=23^2-19cdot 16=225; ,; ; t_{1,2}= frac{23pm 15}{19}$

$a); ; t_1=2; ; to ; ; frac{x}{y} =2; ,; x=2y; ,\\x^2-3xy+y^2=1; ; to ; ; ; (2y)^2-3cdot 2ycdot y+y^2=-1\\4y^2-6y^2+y^2=-1; ,; ; -y^2=-1; ,; ; y^2=1\\y=pm 1; ,; x=pm 2$

$b); ; t_2= frac{8}{19}; ; to ; ; frac{x}{y} = frac{8}{19} ; ,; ; x= frac{8y}{19}\\ frac{64y^2}{361}- frac{3cdot 8y^2}{19} +y^2=-1\\y^2cdot ( frac{64}{361}-frac{24}{19} +1)=-1\\y^2cdot frac{64-24cdot 19+361}{361}=-1\\y^2cdot frac{-31}{361}=-1; ,; ; ; y^2= frac{361}{31} ; ,; ; ; y=pm frac{19}{sqrt{31}} \\x_1=-frac{8cdot 19}{19cdot sqrt{31}}=-frac{8}{sqrt{31}}; ,; ; x_2= frac{8}{sqrt{31}}$

$Otvet:; ; (-1,-1); ,; (1,1); ,; (- frac{8}{sqrt{31}},-frac{19}{sqrt{31}}); ,; (frac{8}{sqrt{31}},frac{19}{sqrt{31}}); .$