Question

4．方程组$\left\{\begin{array}{l}{x+y=12}\\{x+2y=14}\end{array}\right.$的解为（　　）

• A． $\left\{\begin{array}{l}{x=4}\\{y=8}\end{array}\right.$
• B． $\left\{\begin{array}{l}{x=2}\\{y=10}\end{array}\right.$
• C． $\left\{\begin{array}{l}{x=10}\\{y=2}\end{array}\right.$
• D． $\left\{\begin{array}{l}{x=8}\\{y=4}\end{array}\right.$

Answer 1

分析 方程组利用加减消元法求出解即可．

解答 解：$\left\{\begin{array}{l}{x+y=12①}\\{x+2y=14②}\end{array}\right.$，
②-①得：y=2，
把y=2代入①得：x=10，
则方程组的解为$\left\{\begin{array}{l}{x=10}\\{y=2}\end{array}\right.$．
故选C．

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．