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

Решите задачу:

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

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

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

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