题目内容

19.一元一次方程组$\left\{\begin{array}{l}{x+3y=2}\\{x-2y=7}\end{array}\right.$的解的情况是(  )
A.$\left\{\begin{array}{l}{x=5}\\{y=-1}\end{array}\right.$B.$\left\{\begin{array}{l}{x=8}\\{y=-2}\end{array}\right.$C.$\left\{\begin{array}{l}{x=9}\\{y=1}\end{array}\right.$D.$\left\{\begin{array}{l}{x=3}\\{y=-2}\end{array}\right.$

分析 方程组利用加减消元法求出解即可.

解答 解:$\left\{\begin{array}{l}{x+3y=2①}\\{x-2y=7②}\end{array}\right.$,
①-②得:5y=-5,即y=-1,
把y=-1代入①得:x=5,
则方程组的解为$\left\{\begin{array}{l}{x=5}\\{y=-1}\end{array}\right.$,
故选A

点评 此题考查了解二元一次方程组,熟练掌握运算法则是解本题的关键.

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