Question

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

• A． $\left\{\begin{array}{l}{x=5}\\{y=2}\end{array}\right.$
• B． $\left\{\begin{array}{l}{x=4}\\{y=2}\end{array}\right.$
• C． $\left\{\begin{array}{l}{x=5}\\{y=3}\end{array}\right.$
• D． $\left\{\begin{array}{l}{x=-4}\\{y=2}\end{array}\right.$

Answer 1

分析 方程组利用加减消元法求出解即可．

解答 解：$\left\{\begin{array}{l}{x-y=2①}\\{2x+y=10②}\end{array}\right.$，
①+②得：3x=12，
x=4，
把x=4代入①得：y=2，
∴方程组的解为：$\left\{\begin{array}{l}{x=4}\\{y=2}\end{array}\right.$．
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．