Question

20．方程组$\left\{\begin{array}{l}{4x+3y=5}\\{x-2y=4}\end{array}\right.$的解是（　　）

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

Answer 1

分析 方程组利用加减消元法求出解即可．

解答 解：$\left\{\begin{array}{l}{4x+3y=5①}\\{x-2y=4②}\end{array}\right.$，
①-②×4得：11y=-11，即y=-1，
把y=-1代入②得：x=2，
则方程组的解为$\left\{\begin{array}{l}{x=2}\\{y=-1}\end{array}\right.$，
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，消元的方法有：代入消元法与加减消元法．