问题描述: 解方程组:x+2y+3z=62x+3y+z=63x+y+2z=6 1个回答 分类:数学 2014-12-11 问题解答: 我来补答 x+2y+3z=6①2x+3y+z=6②3x+y+2z=6③,①+②+③得6x+6y+6z=18,所以x+y+z=3④,②-①得x+y-2z=0⑤,④-⑤得3z=3,解得z=1,③-①得2x-y-z=0⑥,④+⑥得3x=3,解得x=1,把x=1,z=1代入④得1+y+1=3,解得y=1,所以原方程组的解为x=1y=1z=1. 展开全文阅读