Question

12．二元一次方程组$\left\{\begin{array}{l}{x+3y=2}\\{2x-3y=1}\end{array}\right.$的解是（　　）

• A． $\left\{\begin{array}{l}{x=1}\\{y=\frac{1}{3}}\end{array}\right.$
• B． $\left\{\begin{array}{l}{x=1}\\{y=-\frac{1}{3}}\end{array}\right.$
• C． $\left\{\begin{array}{l}{x=2}\\{y=1}\end{array}\right.$
• D． $\left\{\begin{array}{l}{x=3}\\{y=-\frac{1}{3}}\end{array}\right.$

Answer 1

分析 方程组利用加减消元法求出解即可．

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

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．