Question

Решить систему уравнений

Answer 1

Обозначим x + y = m , xy = n

$\left \{ {{m+n=-15} \atop {m*n=-54}} \right.\\\\\left \{ {{m=-n-15} \atop {(-n-15)*n=-54}} \right.\left \{ {{m=-n-15} \atop {-n^{2}-15n+54=0}} \right.\\\\\left \{ {{m=-n-15} \atop {n^{2}+15n-54=0}} \right.\\\\n_{1}}=3;m_{1}=-3-15=-18\\\\n_{2}=-18;m_{2}=18-15=3\\\\\\\left \{ {{x+y=-18} \atop {xy=3}} \right. \\\\\left \{ {{x=-y-18} \atop {(-y-18)y=3}} \right.\\\\\left \{ {{x=-y-18} \atop {-y^{2}-18y=3}} \right.\\\\\left \{ {{x=-y-18} \atop {y^{2}+18y+3=0}} \right.[tex]</p><p>[tex] y_{1}=\frac{-18+\sqrt{53}}{2}=-9+\sqrt{53}\\\\x_{1}=-(-9+\sqrt{53})-18=9-\sqrt{53} -18=-9-\sqrt{53}\\\\y_{2} =\frac{-18-2\sqrt{53}}{2}=-9-\sqrt{53}\\\\x_{2} =-(-9-\sqrt{53})-18=9+\sqrt{53}-18=-9+\sqrt{53} \\\\\\\left \{ {{x+y=3} \atop {xy=-18}} \right.\\\\\left \{ {{x=3-y} \atop {(3-y)y=-18}} \right. \\\\\left \{ {{x=3-y} \atop {y^{2}-3y-18=0}} \right. \\\\y_{3}=6;x_{3}=3-6=-3\\\\ y_{4}=-3;x_{4}=3-(-3)=3+3=6$

Ответ: $(-9-\sqrt{53};-9+\sqrt{53} ),(-9+\sqrt{53};-9-\sqrt{53}),(-3;6),(6;-3)$