Question

12．已知關(guān)于x，y的方程組$\left\{\begin{array}{l}{x+y=3}\\{ax-by=5}\end{array}\right.$與$\left\{\begin{array}{l}{bx-2ay=1}\\{x-7=y}\end{array}\right.$同解，求$\frac{a}$的值．

Answer 1

分析 根據(jù)題意得出方程組$\left\{\begin{array}{l}{x+y=3}\\{x-7=y}\end{array}\right.$，求出方程組的解，把x、y的值代入方程組$\left\{\begin{array}{l}{ax-by=5}\\{bx-2ay=1}\end{array}\right.$，得出關(guān)于ab的方程組，求出方程組的解即可．

解答 解：∵關(guān)于x，y的方程組$\left\{\begin{array}{l}{x+y=3}\\{ax-by=5}\end{array}\right.$與$\left\{\begin{array}{l}{bx-2ay=1}\\{x-7=y}\end{array}\right.$同解，
∴解方程組$\left\{\begin{array}{l}{x+y=3}\\{x-7=y}\end{array}\right.$，得：$\left\{\begin{array}{l}{x=5}\\{y=-2}\end{array}\right.$，
把x=5，y=-2代入方程組$\left\{\begin{array}{l}{ax-by=5}\\{bx-2ay=1}\end{array}\right.$，得：$\left\{\begin{array}{l}{5a+2b=5}\\{5b+4a=1}\end{array}\right.$，
解得：a=$\frac{23}{17}$，b=-$\frac{4}{5}$．
∴$\frac{a}$=-$\frac{23}{15}$．

點(diǎn)評(píng) 本題考查了解二元一次方程組，二元一次方程組的解的應(yīng)用，關(guān)鍵是能求出關(guān)于a、b的方程組．