Question

10．方程组$\left\{\begin{array}{l}{x=y+2}\\{x+y=4}\end{array}\right.$的解是（　　）

• A． $\left\{\begin{array}{l}{x=1}\\{y=3}\end{array}\right.$
• B． $\left\{\begin{array}{l}{x=3}\\{y=1}\end{array}\right.$
• C． $\left\{\begin{array}{l}{x=2}\\{y=2}\end{array}\right.$
• D． $\left\{\begin{array}{l}{x=2}\\{y=0}\end{array}\right.$

Answer 1

分析 方程组利用代入消元法求出解即可．

解答 解：$\left\{\begin{array}{l}{x=y+2①}\\{x+y=4②}\end{array}\right.$，
把①代入②得：y+2+y=4，
解得：y=1，
把y=1代入①得：x=3，
则方程组的解为$\left\{\begin{array}{l}{x=3}\\{y=1}\end{array}\right.$，
故选B

点评 此题考查了解二元一次方程组，利用了消元的思想，消元的方法有：代入消元法与加减消元法．