Question

4．二元一次方程組$\left\{\begin{array}{l}{2x-y=2}\\{x+y=4}\end{array}\right.$的解是（　　）

• A． $\left\{\begin{array}{l}{x=1}\\{y=6}\end{array}\right.$
• B． $\left\{\begin{array}{l}{x=2}\\{y=2}\end{array}\right.$
• C． $\left\{\begin{array}{l}{x=-3}\\{y=2}\end{array}\right.$
• D． $\left\{\begin{array}{l}{x=3}\\{y=2}\end{array}\right.$

Answer 1

分析 方程組利用加減消元法求出解即可．

解答 解：$\left\{\begin{array}{l}{2x-y=2①}\\{x+y=4②}\end{array}\right.$，
①+②得：3x=6，即x=2，
把x=2代入②得：y=2，
則方程組的解為$\left\{\begin{array}{l}{x=2}\\{y=2}\end{array}\right.$．
故選B．

點評 此題考查了解二元一次方程組，利用了消元的思想，消元的方法有：代入消元法與加減消元法．