已知x+y+8=0,y-3x-2=0,则(x+y+2)*(3x-y+1)=
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已知x+y+8=0,y-3x-2=0,则(x+y+2)*(3x-y+1)=
已知x+y+8=0,y-3x-2=0,则(x+y+2)*(3x-y+1)=
已知x+y+8=0,y-3x-2=0,则(x+y+2)*(3x-y+1)=
x+y+2=x+y+8-6=-6
3x-y+1=-(y-3x-2)-1=-1
所以原式=-6*(-1)=6