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Linear Systemshard
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Consider the system x+y+z=2,x+2y+3z=5,x+3y+4z=8x + y + z = 2, x + 2y + 3z = 5, x + 3y + 4z = 8x+y+z=2,x+2y+3z=5,x+3y+4z=8. How many solutions exist?