①分解（x-y+2)(3x+y-4)=0∴x-y+2=0、3x+y-4=0
②分解（x+y-2)(2x-5y+7)=0∴x+y-2=0、2x-5y+7=0
∴原方程组化为：
x-y+2=0 x-y+2=0 3x+y-4=0 3x+y-4=0
x+y-2=0 2x-5y+7=0 x+y-2=0 2x-5y+7=0
∴原方程组的解是：
x=0 x=-1 x= 1 x=13/17
y=2 y=1 y= 1 y=29/17
3x²-2xy-y²+2x+6y=8,2x²-3xy-5y²+3x+17y=14
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