Question

14．方程组$\left\{\begin{array}{l}{2x+y=3}\\{3x-z=7}\\{x-y+3z=0}\end{array}\right.$的解为（　　）

• A． $\left\{\begin{array}{l}{x=2}\\{y=1}\\{y=-1}\end{array}\right.$
• B． $\left\{\begin{array}{l}{x=2}\\{y=-1}\\{z=1}\end{array}\right.$
• C． $\left\{\begin{array}{l}{x=2}\\{y=-1}\\{z=-1}\end{array}\right.$
• D． $\left\{\begin{array}{l}{x=2}\\{y=1}\\{z=1}\end{array}\right.$

Answer 1

分析 由②③消去z，转化为二元方程组即可解决问题．

解答 解：$\left\{\begin{array}{l}{2x+y=3}&{①}\\{3x-z=7}&{②}\\{x-y+3z=0}&{③}\end{array}\right.$
②×3+③得到：10x-y=21   ④
由①④解得$\left\{\begin{array}{l}{x=2}\\{y=-1}\end{array}\right.$代入②得z=-1，
∴$\left\{\begin{array}{l}{x=2}\\{y=-1}\\{z=-1}\end{array}\right.$，
故选C．

点评 本题考查三元方程组，解题的关键是三元方程组转化为二元方程组，学会转化的数学思想，属于中考常考题型．