Question

11．如果$\left\{\begin{array}{l}{x=-1}\\{y=2}\end{array}\right.$是方程組$\left\{\begin{array}{l}{ax+by=0}\\{bx-cy=1}\end{array}\right.$的解，那么，下列各式中成立的是（　�。�

• A． a+4c=2
• B． 4a+c=2
• C． a+4c+2=0
• D． 4a+c+2=0

Answer 1

分析 把$\left\{\begin{array}{l}{x=-1}\\{y=2}\end{array}\right.$代入方程組$\left\{\begin{array}{l}{ax+by=0}\\{bx-cy=1}\end{array}\right.$可得：$\left\{\begin{array}{l}{-a+2b=0①}\\{-b-2c=1②}\end{array}\right.$，然后②×2+①可得：-a-4c=2，再整理可得答案．

解答 解：把$\left\{\begin{array}{l}{x=-1}\\{y=2}\end{array}\right.$代入方程組$\left\{\begin{array}{l}{ax+by=0}\\{bx-cy=1}\end{array}\right.$可得：$\left\{\begin{array}{l}{-a+2b=0①}\\{-b-2c=1②}\end{array}\right.$，
②×2得：-2b-4c=2③，
①+③得：-a-4c=2，
a+4c+2=0，
故選：C．

點評 此題主要考查了二元一次方程組的解，關鍵是掌握二元一次方程組的兩個方程的公共解，叫做二元一次方程組的解．