xy-2y-x+2=(x-2)y-(x-2)=0,y=1或x=2设y=1,则x^2-3xy+5y-5=x^2-3x+5-5=x^2-3x=x*（x-3）=0所以x=0或x=3设x=2,则x^2-3xy+5y-5=4-6y+5y-5=0,y=1所以有x1=0,y1=1x2=3,y2=1x3=2,y3=1