解：∵y=3xy+x
∴(1-3x)y=x，即y=x/(1-3x)
∴2x+3xy-2y=2x+[3x²/(1-3x)]-[2x/(1-3x)]
=(2x-6x²+3x²-2x)/(1-3x)
=(-3x²)/(1-3x)
x-2xy-y=x-[2x²/(1-3x)]-[x/(1-3x)]
=(x-3x²-2x²-x)/(1-3x)
=(-x²)/(1-3x)
∴(2x+3xy-2y)/(x-2xy-y)
=[(-3x²)/(1-3x)]／[(-x²)/(1-3x)]
=3
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