Question

12．若關(guān)于x，y的方程組$\left\{\begin{array}{l}{5x+3ay=16}\\{-bx+4y=15}\end{array}\right.$（其中a，b是常數(shù)）的解為$\left\{\begin{array}{l}{x=6}\\{y=7}\end{array}\right.$，則方程組 $\left\{\begin{array}{l}{5（x+1）+3a（x-2y）=16}\\{-b（x+1）+4（x-2y）=15}\end{array}\right.$的解為（　�。�

• A． $\left\{\begin{array}{l}{x=6}\\{y=7}\end{array}\right.$
• B． $\left\{\begin{array}{l}{x=5}\\{y=-1}\end{array}\right.$
• C． $\left\{\begin{array}{l}{x=5}\\{y=1}\end{array}\right.$
• D． $\left\{\begin{array}{l}{x=5.5}\\{y=-1}\end{array}\right.$

Answer 1

分析 根據(jù)題意得到關(guān)于x，y的二元一次方程組$\left\{\begin{array}{l}{x+1=6①}\\{x-2y=7②}\end{array}\right.$，解方程組即可求解．

解答 解：依題意有$\left\{\begin{array}{l}{x+1=6①}\\{x-2y=7②}\end{array}\right.$，
解①得x=5，
把x=5代入②得5-2y=7，解得y=-1．
故方程組 $\left\{\begin{array}{l}{5（x+1）+3a（x-2y）=16}\\{-b（x+1）+4（x-2y）=15}\end{array}\right.$的解為$\left\{\begin{array}{l}{x=5}\\{y=-1}\end{array}\right.$．
故選：B．

點評 本題考查了二元一次方程組的解，能使方程組中每個方程的左右兩邊相等的未知數(shù)的值即是方程組的解．解題的關(guān)鍵是要知道兩個方程組之間解的關(guān)系．