230问答网 > {x=3①,2x+y-3z=11②,x+y+z=12③ 这个三元一次方程组怎么解？

{x=3①,2x+y-3z=11②,x+y+z=12③ 这个三元一次方程组怎么解？

x=3 ① 2x+y-3z=11 ② x+y+z=12 ③ 这个三元一次方程组怎么解？
2025-03-07 03:49:12
推荐回答（2个）
回答1：

代入X=3
第一条方程式等于：6+y-3z=11也就是y-3z=5
第二条方程式等于：3+y+z=12也就是y+z=9于是y=9-z
代入第一条得：9-z-3z=5
移项解得z=1 所以y=9-1=8

回答2：

三元方程联立求解： 由1可知X=3.
代入2.3式。得：6+y-3z =11, 3+y+z =12.
联立求解得：y=8, z=1.
方程组的解为X=3, Y=8, Z=1

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