Question

7．方程组$\left\{\begin{array}{l}{x+2y=4}\\{x=y+1}\end{array}\right.$的解为（　　）

• A． $\left\{\begin{array}{l}{x=4}\\{y=3}\end{array}\right.$
• B． $\left\{\begin{array}{l}{x=2}\\{y=1}\end{array}\right.$
• C． $\left\{\begin{array}{l}{x=1}\\{y=0}\end{array}\right.$
• D． $\left\{\begin{array}{l}{x=-1}\\{y=-2}\end{array}\right.$

Answer 1

分析 把x的值代入第一个方程可以求得y的值；然后求x的值．

解答 解：$\left\{\begin{array}{l}{x+2y=4①}\\{x=y+1②}\end{array}\right.$，
把②代入①得到：y+1+2y=4，
解得y=1，
把③代入②得到：x+1+1=2，
则原方程组的解为：$\left\{\begin{array}{l}{x=2}\\{y=1}\end{array}\right.$．
故选：B．

点评 本题考查了解二元一次方程组．这类题目的解题关键是掌握方程组解法中的加减消元法和代入法．