方程组｛100x+10y+z=48(x+y+z) ｛x+z=y+3 ｛100x+10y+z=100z+10y+x+198 怎么解
问题描述：
方程组｛100x+10y+z=48(x+y+z) ｛x+z=y+3 ｛100x+10y+z=100z+10y+x+198 怎么解
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)