Question

11．方程组$\left\{\begin{array}{l}{x+3y=5}\\{2x-3y=1}\end{array}\right.$的解是（　　）

• A． $\left\{\begin{array}{l}{x=-1}\\{y=-1}\end{array}\right.$
• B． $\left\{\begin{array}{l}{x=2}\\{y=1}\end{array}\right.$
• C． $\left\{\begin{array}{l}{x=-2}\\{y=2}\end{array}\right.$
• D． $\left\{\begin{array}{l}{x=-2}\\{y=-1}\end{array}\right.$

Answer 1

分析 用加减消元法解方程组，即可解答．

解答 解：$\left\{\begin{array}{l}{x+3y=5①}\\{2x-3y=1②}\end{array}\right.$
①+②得：3x=6，
解得：x=2，
把x=2代入①得：2+3y=5，
解得：y=1，
∴方程组的解为：$\left\{\begin{array}{l}{x=2}\\{y=1}\end{array}\right.$
故选：B．

点评 本题考查了解二元一次方程组，解决本题的关键是用加减消元法解方程组．