已知3x+2y=4+z,2x+2z=6+y,问是否存在x、y、z的正整数值,使得x+y+z
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已知3x+2y=4+z,2x+2z=6+y,问是否存在x、y、z的正整数值,使得x+y+z
假设x+y+z z=3x+2y-4
2x+2z=6+y ===> y=2x+2z-6=2x-6+6x+4y-8=8x-14+4y===> 3y=14-8x
z=3x+2y-4 ===> 3z=9x+6y-12=9x-12+28-16x=16-7x
3x+3y+3z=3x+14-8x+16-7x=30-12x
x+y+z=10-4x
10-4x3(成立)
x=1时,3y=14-8x ===> y=(14-8)/3=2(成立)
z=3x+2y-4=3+2*2-4=3(成立)
x=2时,3y=14-8x ===> y=(14-8*2)/3=-2/3(不成立)
因此,x=1、y=2、z=3满足题意.
2x+2z=6+y ===> y=2x+2z-6=2x-6+6x+4y-8=8x-14+4y===> 3y=14-8x
z=3x+2y-4 ===> 3z=9x+6y-12=9x-12+28-16x=16-7x
3x+3y+3z=3x+14-8x+16-7x=30-12x
x+y+z=10-4x
10-4x3(成立)
x=1时,3y=14-8x ===> y=(14-8)/3=2(成立)
z=3x+2y-4=3+2*2-4=3(成立)
x=2时,3y=14-8x ===> y=(14-8*2)/3=-2/3(不成立)
因此,x=1、y=2、z=3满足题意.
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