Question

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． $\left\{\begin{array}{l}{x=3}\\{y=-2}\end{array}\right.$

Answer 1

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．