Question

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

• A． $\left\{\begin{array}{l}{x=1}\\{y=-1}\end{array}\right.$
• B． $\left\{\begin{array}{l}{x=3}\\{y=1}\end{array}\right.$
• C． $\left\{\begin{array}{l}{x=0}\\{y=1}\end{array}\right.$
• D． $\left\{\begin{array}{l}{x=3}\\{y=-1}\end{array}\right.$

Answer 1

分析 方程組利用加減消元法求出解即可．

解答 解：$\left\{\begin{array}{l}{x-y=2①}\\{2x+y=1②}\end{array}\right.$，
①+②得：3x=3，即x=1，
把x=1代入①得：y=-1，
則方程組的解為$\left\{\begin{array}{l}{x=1}\\{y=-1}\end{array}\right.$，
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．