Question

2．方程组$\left\{\begin{array}{l}{x-y=1}\\{x+y=3}\end{array}\right.$的解是（　　）

• A． $\left\{\begin{array}{l}{x=1}\\{y=2}\end{array}\right.$
• B． $\left\{\begin{array}{l}{x=1}\\{y=3}\end{array}\right.$
• C． $\left\{\begin{array}{l}{x=3}\\{y=1}\end{array}\right.$
• D． $\left\{\begin{array}{l}{x=2}\\{y=1}\end{array}\right.$

Answer 1

分析 利用加减消元法求出方程组的解，即可作出判断．

解答 解：$\left\{\begin{array}{l}{x-y=1①}\\{x+y=3②}\end{array}\right.$，
①+②得：2x=4，即x=2，
把x=2代入①得：y=1，
则方程组的解为$\left\{\begin{array}{l}{x=2}\\{y=1}\end{array}\right.$，
故选D

点评 此题考查了二元一次方程组的解，求出方程组的解是解本题的关键．