Question

14．二元一次方程组$\left\{\begin{array}{l}{3x+2y=6}\\{y=x-2}\end{array}\right.$的解是（　　）

• A． $\left\{\begin{array}{l}{x=\frac{1}{2}}\\{y=-\frac{3}{2}}\end{array}\right.$
• B． $\left\{\begin{array}{l}{x=2}\\{y=0}\end{array}\right.$
• C． $\left\{\begin{array}{l}{x=-\frac{3}{2}}\\{y=\frac{1}{2}}\end{array}\right.$
• D． $\left\{\begin{array}{l}{x=0}\\{y=-2}\end{array}\right.$

Answer 1

分析 方程组利用代入消元法求出解即可．

解答 解：$\left\{\begin{array}{l}{3x+2y=6①}\\{y=x-2②}\end{array}\right.$，
把②代入①得：3x+2x-4=6，
解得：x=2，
把x=2代入②得：y=0，
则方程组的解为$\left\{\begin{array}{l}{x=2}\\{y=0}\end{array}\right.$，
故选B．

点评 此题考查了解二元一次方程组，利用了消元的思想，消元的方法为：加减消元法与代入消元法．