Question

ПОМОГИТЕ!!!!!!!!!!решите систему уравнений методом сложения!!!
х^2+4y^2=85
xy=-21

Answer 1

$\left \{ {{x^{2} +4y^{2}=85 } \atop {xy=-21} \ |\cdot 4} \right. \\\\\\+\left \{ {{x^{2} +4y^{2}=85 } \atop {4xy=-84}} \right.\\-------\\x^{2}+4xy+4y^{2}=1\\\\(x+2y)^{2}=1\\\\\\\left[\begin{array}{ccc}x+2y=1\\x+2y=-1\end{array}\right\\\\\\1) \\\left \{ {{x+2y=1} \atop {xy=-21}} \right. \\\\\\\left \{ {{x=1-2y} \atop {y\cdot(1-2y)=-21}} \right.\\\\\\\left \{ {{x=1-2y} \atop {y-2y^{2}+21=0 }} \right. \\\\\\\left \{ {{x=1-2y} \atop {2y^{2}-y-21=0 }} \right.$

$\left[\begin{array}{ccc}\left \{ {{x=1-2y} \atop {y_{1} =3,5}} \right. \\ \left \{ {{x=1-2y} \atop {y_{2} =-3}} \right. \end{array}\right\\\\\\\left[\begin{array}{ccc}\left \{ {{x_{1} =-6} \atop {y_{1}=3,5 }} \right. \\\left \{ {{x_{2} =7} \atop {y_{2}=-3 }} \right. \end{array}\right$

$2) \\\left \{ {{x+2y=-1} \atop {xy=-21}} \right. \\\\\\\left \{ {{x=-1-2y} \atop {y\cdot(-1-2y)=-21}} \right.\\\\\\\left \{ {{x=-1-2y} \atop {-y-2y^{2}+21=0 }} \right. \\\\\\\left \{ {{x=-1-2y} \atop {2y^{2}+y-21=0 }} \right.$

$\left[\begin{array}{ccc}\left \{ {{x=-1-2y} \atop {y_{1} =-3,5}} \right. \\ \left \{ {{x=-1-2y} \atop {y_{2} =3}} \right. \end{array}\right\\\\\\\left[\begin{array}{ccc}\left \{ {{x_{1} =6} \atop {y_{1}=-3,5 }} \right. \\\left \{ {{x_{2} =-7} \atop {y_{2}=3 }} \right. \end{array}\right\\\\Otvet:\boxed{(-6 \ ; \ 3,5) \ , \ (7 \ ; \ -3) \ , \ (6 \ ; \ -3,5) \ , \ (-7 \ ; \ 3)}$