Question

7．$\left\{\begin{array}{l}{x=-2}\\{y=2}\end{array}\right.$是下列哪个二元一次方程组的解（　　）

• A． $\left\{\begin{array}{l}{2x-y=-3}\\{4x+y=6}\end{array}\right.$
• B． $\left\{\begin{array}{l}{2x-y=-6}\\{2x+y=2}\end{array}\right.$
• C． $\left\{\begin{array}{l}{x+y=0}\\{x-y=2}\end{array}\right.$
• D． $\left\{\begin{array}{l}{2x+y=-2}\\{x+2y=2}\end{array}\right.$

Answer 1

分析 根据加减消元法，求出二元一次方程组的解，即可解答．

解答 解：A、方程组$\left\{\begin{array}{l}{2x-y=-3}\\{4x+y=6}\end{array}\right.$的解为$\left\{\begin{array}{l}{x=\frac{1}{2}}\\{y=4}\end{array}\right.$，故不合题意；
B、方程组$\left\{\begin{array}{l}{2x-y=-6}\\{2x+y=2}\end{array}\right.$的解为$\left\{\begin{array}{l}{x=-1}\\{y=4}\end{array}\right.$，故不合题意；
C、方程组$\left\{\begin{array}{l}{x+y=0}\\{x-y=2}\end{array}\right.$的解为$\left\{\begin{array}{l}{x=1}\\{y=-1}\end{array}\right.$，故不合题意；
D、方程组$\left\{\begin{array}{l}{2x+y=-2}\\{x+2y=2}\end{array}\right.$的解为$\left\{\begin{array}{l}{x=-2}\\{y=2}\end{array}\right.$，正确；
故选：D．

点评 本题考查了二元一次方程组的解，解决本题的关键是用加减法解二元一次方程组．