Question

2．方程組$\left\{\begin{array}{l}{x+y=5}\\{x-2y=2}\end{array}\right.$的解是（　　）

• A． $\left\{\begin{array}{l}{x=1}\\{y=4}\end{array}\right.$
• B． $\left\{\begin{array}{l}{x=3}\\{y=2}\end{array}\right.$
• C． $\left\{\begin{array}{l}{x=8}\\{y=-3}\end{array}\right.$
• D． $\left\{\begin{array}{l}{x=4}\\{y=1}\end{array}\right.$

Answer 1

分析 方程組利用加減消元法求出解即可．

解答 解：$\left\{\begin{array}{l}{x+y=5①}\\{x-2y=2②}\end{array}\right.$，
①-②得：3y=3，即y=1，
把y=1代入①得：x=4，
則方程組的解為$\left\{\begin{array}{l}{x=4}\\{y=1}\end{array}\right.$，
故選D．

點(diǎn)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．