Question

x+y=3,    x^2+2xy+2y^2=18

Answer 1

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

 

2-й случай

 

 

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

 

1-й случай

 

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

 

2-й случай

 

 

$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

 

 

[tex]left { {{x=3-y} atop {y=3}} right. \ \left { {{x=0} atop {y=3}} right." />

 

Ответ: (x=6 y=-3) (x=0 y=3)