题目内容

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

分析 根据y的系数互为相反数,利用加减消元法求解即可.

解答 解:$\left\{\begin{array}{l}{x-y=1①}\\{2x+y=5②}\end{array}\right.$,
①+②得,3x=6,
解得x=2,
把x=2代入①得,2-y=1,
解得y=1,
所以方程组的解是$\left\{\begin{array}{l}{x=2}\\{y=1}\end{array}\right.$,
故选D.

点评 本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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