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方程组{100x+10y+z=48(x+y+z) {x+z=y+3 {100x+10y+z=100z+10y+x+198 怎么解
人气:207 ℃ 时间:2020-06-11 17:03:08
解答
100x+10y+z=48(x+y+z).(1)
x+z=y+3.(2)
100x+10y+z=100z+10y+x+198,99x=99z+198,x=z+2.(3)
将③代入②:
y=2z-1
均代入①:
52x=38y+47z
52z+104=76z-38+47z
71z=142
z=2
代入得y=2z-1=3
x=y+3-z=4
∴(x,y,z)=(4,3,2)
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