Question

11．方程組$\left\{\begin{array}{l}{x-y=3}\\{3x-y=1}\end{array}\right.$的解是（　�。�

• A． $\left\{\begin{array}{l}{x=-1}\\{y=-4}\end{array}\right.$
• B． $\left\{\begin{array}{l}{x=1}\\{y=-4}\end{array}\right.$
• C． $\left\{\begin{array}{l}{x=-1}\\{y=4}\end{array}\right.$
• D． $\left\{\begin{array}{l}{x=1}\\{y=4}\end{array}\right.$

Answer 1

分析 利用加減消元法求出方程組的解即可．

解答 解：$\left\{\begin{array}{l}{x-y=3①}\\{3x-y=1②}\end{array}\right.$，
①-②得：-2x=2，
解得：x=-1，
把x=-1代入①得：y=-4，
則方程組的解為$\left\{\begin{array}{l}{x=-1}\\{y=-4}\end{array}\right.$，
故選A

點評 此題考查了解二元一次方程組，利用了消元的思想，消元的方法有：代入消元法與加減消元法．