Question

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

• A． $\left\{\begin{array}{l}{x=1}\\{y=2}\end{array}\right.$
• B． $\left\{\begin{array}{l}{x=-1}\\{y=2}\end{array}\right.$
• 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 方程組利用加減消元法求出解即可．

解答 解：$\left\{\begin{array}{l}{2x-y=5①}\\{3x+4y=2②}\end{array}\right.$，
①×4+②得：11x=22，即x=2，
把x=2代入①得：y=-1，
則方程組的解為$\left\{\begin{array}{l}{x=2}\\{y=-1}\end{array}\right.$，
故選C

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