题目内容

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.$\left\{\begin{array}{l}{x=7}\\{y=2}\end{array}\right.$

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.

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