Question

12．已知方程組$\left\{\begin{array}{l}{5x+y=3}\\{ax+5y=4}\end{array}\right.$和$\left\{\begin{array}{l}{x-2y=5}\\{5x+by=1}\end{array}\right.$有相同的解，則a，b的值為（　�。�

• A． $\left\{\begin{array}{l}{a=1}\\{b=2}\end{array}\right.$
• B． $\left\{\begin{array}{l}{a=-4}\\{b=-6}\end{array}\right.$
• C． $\left\{\begin{array}{l}{a=-6}\\{b=2}\end{array}\right.$
• D． $\left\{\begin{array}{l}{a=14}\\{b=2}\end{array}\right.$

Answer 1

分析 因為方程組$\left\{\begin{array}{l}{5x+y=3}\\{ax+5y=4}\end{array}\right.$和$\left\{\begin{array}{l}{x-2y=5}\\{5x+by=1}\end{array}\right.$有相同的解，所以把5x+y=3和x-2y=5聯(lián)立解之求出x、y，再代入其他兩個方程即可得到關(guān)于a、b的方程組，解方程組即可求解．

解答 解：∵方程組$\left\{\begin{array}{l}{5x+y=3}\\{ax+5y=4}\end{array}\right.$和$\left\{\begin{array}{l}{x-2y=5}\\{5x+by=1}\end{array}\right.$有相同的解，
∴方程組$\left\{\begin{array}{l}{5x+y=3}\\{x-2y=5}\end{array}\right.$的解也它們的解，
解得：$\left\{\begin{array}{l}{x=1}\\{y=-2}\end{array}\right.$，
代入其他兩個方程得$\left\{\begin{array}{l}{a-10=4}\\{5-2b=1}\end{array}\right.$，
解得：$\left\{\begin{array}{l}{a=14}\\{b=2}\end{array}\right.$，
故選D．

點評 本題主要考查了二元一次方程的解及二元一次方程組的解法，正確理解題意，然后根據(jù)題意得到關(guān)于待定系數(shù)的方程組，解方程組是解答此題的關(guān)鍵．