Question

8．已知關(guān)于x，y的方程組$\left\{\begin{array}{l}{x+y=2}\\{ax+by=1}\end{array}\right.$與$\left\{\begin{array}{l}{x-2y=5}\\{ax-by=4}\end{array}\right.$的解相同，則a+b=$\frac{7}{3}$．

Answer 1

分析 由于方程組$\left\{\begin{array}{l}{x+y=2}\\{ax+by=1}\end{array}\right.$與$\left\{\begin{array}{l}{x-2y=5}\\{ax-by=4}\end{array}\right.$的解相同，所以把x+y=2和x-2y=5聯(lián)立解之求出x、y，再代入其他兩個(gè)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$\left\{\begin{array}{l}{x+y=2}\\{ax+by=1}\end{array}\right.$與$\left\{\begin{array}{l}{x-2y=5}\\{ax-by=4}\end{array}\right.$的解相同，
∴方程組$\left\{\begin{array}{l}{x+y=2}\\{x-2y=5}\end{array}\right.$的解也它們的解，
解之得：$\left\{\begin{array}{l}{x=3}\\{y=-1}\end{array}\right.$，
代入其他兩個(gè)方程得$\left\{\begin{array}{l}{3a-b=1}\\{3a+b=4}\end{array}\right.$，
解之得：$\left\{\begin{array}{l}{a=\frac{5}{6}}\\{b=\frac{3}{2}}\end{array}\right.$，
∴a+b=$\frac{5}{6}$+$\frac{3}{2}$=$\frac{7}{3}$．
故答案為$\frac{7}{3}$．

點(diǎn)評(píng) 此題主要考查了二元一次方程組的解及二元一次方程組的解法，解題時(shí)首先正確理解題意，然后根據(jù)題意得到關(guān)于待定系數(shù)的方程組，解方程組即可求解．