若x^2+xy+y=14,y^2+xy+x=28,求x+y的值.

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若x^2+xy+y=14,y^2+xy+x=28,求x+y的值.
若x^2+xy+y=14,y^2+xy+x=28,求x+y的值.

若x^2+xy+y=14,y^2+xy+x=28,求x+y的值.
x^2+xy+y=14,y^2+xy+x=28
两式相加得
x²+2xy+y²+(x+y)=42
(x+y)²+(x+y)-42=0
(x+y+7)(x+y-6)=0
所以x+y+7=0或x+y-6=0
即x+y=-7或x+y=6

x^2+xy+y=14 ①
y^2+xy+x=28 ②
①+②得,x^2+xy+y+y^2+xy+x=42
(x^2+2xy+y^2)+(x+y)=42
(x+y)^2+(x+y+1)=42
(x+y)(x+y+1)=42
因为42=6*7=(-6)*(-7)
所以x+y=6或-7

X+Y=6或7
两式子相加
可得（X+Y)^2+(X+Y)-42=0
因式分解得
(X+Y-6)(X+Y+7)=0
X+Y=6或7

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