Question

9．方程组$\left\{\begin{array}{l}{x+y=10}\\{2x+y=16}\end{array}\right.$的解是（　　）

• A． $\left\{\begin{array}{l}{x=5}\\{y=5}\end{array}\right.$
• B． $\left\{\begin{array}{l}{x=6}\\{y=4}\end{array}\right.$
• C． $\left\{\begin{array}{l}{x=2}\\{y=8}\end{array}\right.$
• D． $\left\{\begin{array}{l}{x=7}\\{y=2}\end{array}\right.$

Answer 1

分析 方程组利用加减消元法求出解即可．

解答 解：$\left\{\begin{array}{l}{x+y=10①}\\{2x+y=16②}\end{array}\right.$，
②-①得：x=6，
把x=6代入①得：y=4，
则方程组的解为$\left\{\begin{array}{l}{x=6}\\{y=4}\end{array}\right.$，
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．