题目内容

2.方程组$\left\{\begin{array}{l}{x+y=5}\\{x-2y=2}\end{array}\right.$的解是(  )
A.$\left\{\begin{array}{l}{x=1}\\{y=4}\end{array}\right.$B.$\left\{\begin{array}{l}{x=3}\\{y=2}\end{array}\right.$C.$\left\{\begin{array}{l}{x=8}\\{y=-3}\end{array}\right.$D.$\left\{\begin{array}{l}{x=4}\\{y=1}\end{array}\right.$

分析 方程组利用加减消元法求出解即可.

解答 解:$\left\{\begin{array}{l}{x+y=5①}\\{x-2y=2②}\end{array}\right.$,
①-②得:3y=3,即y=1,
把y=1代入①得:x=4,
则方程组的解为$\left\{\begin{array}{l}{x=4}\\{y=1}\end{array}\right.$,
故选D.

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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