Question

6．方程组$\left\{\begin{array}{l}{x-z=4}\\{z-2y=-1}\\{x+y-z=-1}\end{array}\right.$的解是（　　）

• A． $\left\{\begin{array}{l}{x=7}\\{y=-5}\\{z=-11}\end{array}\right.$
• B． $\left\{\begin{array}{l}{x=-7}\\{y=5}\\{z=-11}\end{array}\right.$
• C． $\left\{\begin{array}{l}{x=-7}\\{y=-5}\\{z=-11}\end{array}\right.$
• D． $\left\{\begin{array}{l}{x=7}\\{y=-5}\\{z=11}\end{array}\right.$

Answer 1

分析 方程组利用加减消元法求出解即可．

解答 解：$\left\{\begin{array}{l}{x-z=4①}\\{z-2y=-1②}\\{x+y-z=-1③}\end{array}\right.$，
③-①得：y=-5，
把y=-5代入②得：z=-11，
把z=-11代入①得：x=-7，
则方程组的解为$\left\{\begin{array}{l}{x=-7}\\{y=-5}\\{z=-11}\end{array}\right.$，
故选C．

点评 此题考查了解三元一次方程组，利用了消元的思想，消元的方法有：代入消元法与加减消元法．