Question

14．方程組$\left\{\begin{array}{l}{2x+y=3}\\{3x-z=7}\\{x-y+3z=0}\end{array}\right.$的解為（　�。�

• 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，轉(zhuǎn)化為二元方程組即可解決問題．

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，
∴$\left\{\begin{array}{l}{x=2}\\{y=-1}\\{z=-1}\end{array}\right.$，
故選C．

點評 本題考查三元方程組，解題的關(guān)鍵是三元方程組轉(zhuǎn)化為二元方程組，學會轉(zhuǎn)化的數(shù)學思想，屬于中考�？碱}型．