Решить систему уравнений xy=-10 x^2+y^2=29 В ответе должно получится...

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$\left \{ {{xy=-10} \atop {x^2+y^2=29}} \right. \; \left \{ {{2xy=-20} \atop {x^2+2xy+y^2=29-20}} \right. \; \left \{ {{xy=-10} \atop {(x+y)^2=9}} \right. \; \left \{ {{xy=-10} \atop {x+y=\pm 3}} \right. \\\\ a)\; \; \left \{ {{xy=-10} \atop {x+y=3}} \right. \; \; \; \; \; b)\; \; \left \{ {{xy=-10} \atop {x+y=-3}} \right. \\\\ \left \{ {{x(3-x)=-10} \atop {y=3-x}} \right. \; \; \; \; \; \; \; \; \; \left \{ {{x(-3-x)=-10} \atop {y=-3-x}} \right.$

$\left \{ {{x^2-3x-10=0} \atop {y=3-x}} \right. \; \; \; \; \; \; \; \; \; \left \{ {{x^2+3x-10=0} \atop {y=-3-x}} \right. \\\\ \left \{ {{x_1=-2,\; x_2=5} \atop {y_1=5,\; \; y_2=-2}} \right. \; \; \; \; \; \; \; \; \; \left \{ {{x_3=-5,\; x_4=2} \atop {y_3=2,\; \; y_4=-5}} \right. \\\\Otvet:\; \; (-2,5)\; ,\; (5,-2)\; ,\; (-5,2)\; ,\; (2,-5)\; .$