Question

Решить систему: x^2+y^2-7xy=55, { 9xy+x+y=-55;

Answer 1

$+\left \{ {{x^2+y^2-7xy=55} \atop {9xy+x+y=-55}} \right. \\ x^2+2xy+y^2+x+y=0 \\ (x+y)^2+(x+y)=0 \\ (x+y)(x+y+1)=0 \\ \left \{ {{x+y=0} \atop {9xy+x+y=-55}} \right. \left \{ {{x+y+1=0} \atop {9xy+x+y=-55}} \right. \\ \left \{ {{x=-y} \atop {-9y^2=-55}} \right. \left \{ {{x=-y-1} \atop {-9y(y+1)-1=-55}} \right. \\ \left \{ {{x=-y} \atop {y^2= \frac{55}{9} }} \right. \left \{ {{x=-y-1} \atop {-9y^2-9y+54=0}} \right.$
$\left \{ {{x=\mp \frac{ \sqrt{55} }{3} } \atop {y=\pm \frac{ \sqrt{55} }{3} }} \right. \left \{ {{x=2} \atop {y=-3}} \right. \left \{ {{x=-3} \atop {y=2}} \right.$

ответ: $(- \frac{ \sqrt{55} }{3}; \frac{ \sqrt{55} }{3}), (\frac{ \sqrt{55} }{3}; -\frac{ \sqrt{55} }{3}), (-3; 2), (2; -3)$