由 |x+y+1|+(x+y^2)=0
可得
|x+y+1|=0
x+y^2=0
又可得
x+y+1=0 或 -x-y-1=0也即有x+y+1=0
x+y^2=0
即有
x+y+1=x+y^2
y^2-y-1=0
解之 y=(1±√5)/2
在由x+y+1=0得
(1) 当y=(1+√5)/2
x= -(3+√5)/2
(2)当y=(1-√5)/2
x=-(3-√5)/2
则（1）
x+2y= -(3+√5)/2+2*(1+√5)/2=(√5-1)/2
(2)
x+2y= -(3-√5)/2+2*(1-√5)/2= -(√5+1)/2