Question

9．二元一次方程組$\left\{\begin{array}{l}{2x+y=10}\\{x=2y}\end{array}\right.$的解是（　　）

• A． $\left\{\begin{array}{l}{x=4}\\{y=3}\end{array}\right.$
• B． $\left\{\begin{array}{l}{x=6}\\{y=3}\end{array}\right.$
• C． $\left\{\begin{array}{l}{x=4}\\{y=2}\end{array}\right.$
• D． $\left\{\begin{array}{l}{x=2}\\{y=4}\end{array}\right.$

Answer 1

分析 方程組利用代入消元法求出解即可．

解答 解：$\left\{\begin{array}{l}{2x+y=10①}\\{x=2y②}\end{array}\right.$，
把②代入①得：4y+y=10，
解得：y=2，
把y=2代入②得：x=4，
則方程組的解為$\left\{\begin{array}{l}{x=4}\\{y=2}\end{array}\right.$，
故選C

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