решите систему уравнений 2(x+y)^2

onepeace

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пусть (x+y)=a, тогда:

2a^2-7a+3=0

D=49-24=25

a1=(7+5)/4=3 a2=(7-5)/4=0.5

a=3 или a=0.5

делаем обратную подстановку:

х+у=3 или х+у=0.5

$\left \{ {\left \{ {{x+y=3} \atop {2x-3y=-1}} \right.} \atop { \left \{ {{x+y=0.5} \atop {2x-3y=-1}} \right.}} \right.$ $\left \{ {\left \{ {{x=3-y} \atop {6-2y-3y=-1}} \right.} \atop { \left \{ {{x=0.5-y} \atop {1-2y-3y=-1}} \right.}} \right.$ $\left \{ {\left \{ {{x=2} \atop {y=1}} \right.} \atop { \left \{ {{x=0.1} \atop {y=0.4}} \right.}} \right.$

48 votes Thanks 89

Andrew542

Verified answer

(x+y-3)*(2x+2y-1)=0
2x-3y=-1

x+y-3=0
2x+2y-1=0
2x-3y=-1

1 случай:
x+y-3=0
2x-3y=-1

x=-y+3
2x-3y=-1

x=-y+3
2(-y+3)-3y=-1

x=-y+3
y=1,4

x=-1,4+3
y=1,4

x=1,6
y=1,4

2 случай:

2x+2y-1=0
2x-3y=-1

x=-2y+1 / 2
2x-3y=-1

x=-2y-1 / 2
2(-2y-1 / 2)-3y=-1

x=-2y-1 / 2
-2*2y-1 / 2 - 3y=-1

x=-2y-1 / 2
y=0,4

x=- 2*0,4-1 / 2
y=0,4

x=0,1
y=0,4

Окончательный ответ:
x=1,6
y=1,4

x=0,1
y=0,4

11 votes Thanks 25

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