Ответы и объяснения

Лучший Ответ!
2012-12-10T20:21:04+04:00

\left \{ {{x+y=3} \atop {x^{2}+2xy+2y^{2}=18}} \right. \\ \\\left \{ {{x=3-y} \atop {x^{2}+2xy+2y^{2}=18}} \right. \\ \\\left \{ {{x=3-y} \atop {(3-y)^{2}+2(3-y)y+2y^{2}=18}} \right. \\ \\(3-y)^{2}+2(3-y)y+2y^{2}=18 \\ \\y^{2}-6y+9-2y^{2}+6y+2y^{2}-18=0 \\ \\y^{2}-9=0 \\ \\y^{2}=9 \\ \\ y=-3; y=3 \\  

 

1-й случай

 

\left \{ {{x=3-(-3))} \atop {y=-3}} \right. \\ \\\left \{ {{x=6} \atop {y=-3}} \right. \\

 

2-й случай

 

 

\left \{ {{x=3-y} \atop {y=3}} \right. \\ \\\left \{ {{x=0} \atop {y=3}} \right.

 

Ответ: (x=6 y=-3) (x=0 y=3)