Question

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

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

Answer 1

分析 方程组利用代入消元法求出解即可．

解答 解：$\left\{\begin{array}{l}{x+2y=10①}\\{y=2x②}\end{array}\right.$，
把②代入①得：x+4x=10，即x=2，
把x=2代入②得：y=4，
则方程组的解为$\left\{\begin{array}{l}{x=2}\\{y=4}\end{array}\right.$．
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．