题目内容

12.方程组$\left\{\begin{array}{l}{x-y=2}\\{2x+y=10}\end{array}\right.$的解是(  )
A.$\left\{\begin{array}{l}{x=5}\\{y=2}\end{array}\right.$B.$\left\{\begin{array}{l}{x=4}\\{y=2}\end{array}\right.$C.$\left\{\begin{array}{l}{x=5}\\{y=3}\end{array}\right.$D.$\left\{\begin{array}{l}{x=-4}\\{y=2}\end{array}\right.$

分析 方程组利用加减消元法求出解即可.

解答 解:$\left\{\begin{array}{l}{x-y=2①}\\{2x+y=10②}\end{array}\right.$,
①+②得:3x=12,
x=4,
把x=4代入①得:y=2,
∴方程组的解为:$\left\{\begin{array}{l}{x=4}\\{y=2}\end{array}\right.$.
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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