解方程组{3x+y-z=4,2x-y+3z=12,x+y+z=6}

问题描述:

解方程组{3x+y-z=4,2x-y+3z=12,x+y+z=6}

{3x+y-z=4①,2x-y+3z=12②,x+y+z=6③}
①+②得5x+2z=16④,②+③得3x+4z=18⑤
④×2—⑤得7x=14,x=2
所以z=3、y=1所以方程组的解为x=2、y=1、z=3