Question

7．方程组$\left\{\begin{array}{l}{x-1=0}\\{x+1=y}\end{array}\right.$的解是（　　）

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

Answer 1

分析 利用代入消元法求出方程组的解，即可作出判断．

解答 解：$\left\{\begin{array}{l}{x-1=0①}\\{x+1=y②}\end{array}\right.$，
由①得：x=1，
把x=1代入②得：y=2，
则方程组的解为$\left\{\begin{array}{l}{x=1}\\{y=2}\end{array}\right.$，
故选A

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．