题目二元一次方程组{0.2x−0.3y=3x3+y=2\left\{\begin{array}{l}0.2x-0.3y=3\\ \dfrac{x}{3}+y=2\\\end{array}\right.{0.2x−0.3y=33x+y=2，的解是A.{x=9,y=−1;\left\{\begin{array}{l}x=9,\\ y=-1;\\\end{array}\right.{x=9,y=−1;B.{x=8,y=−4;\left\{\begin{array}{l}x=8,\\ y=-4;\\\end{array}\right.{x=8,y=−4;C.{x=12,y=2;\left\{\begin{array}{l}x=12,\\ y=2;\\\end{array}\right.{x=12,y=2;D.{x=12,y=−2.\left\{\begin{array}{l}x=12,\\ y=-2.\\\end{array}\right.{x=12,y=−2.知识点：解二元一次方程组——加减消元法章节：方程与不等式 / 二元一次方程组
