Question

13．若关于m，n的二元一次方程组$\left\{\begin{array}{l}{3m-an=16}\\{2m-bn=15}\end{array}\right.$的解是$\left\{\begin{array}{l}{m=7}\\{n=1}\end{array}\right.$，那么关于x，y的二元一次方程组$\left\{\begin{array}{l}{3（x+y）-a（x-y）=16}\\{2（x+y）-b（x-y）=15}\end{array}\right.$的解$\left\{\begin{array}{l}{x=4}\\{y=3}\end{array}\right.$．

Answer 1

分析 根据方程组$\left\{\begin{array}{l}{3m-an=16}\\{2m-bn=15}\end{array}\right.$的解是$\left\{\begin{array}{l}{m=7}\\{n=1}\end{array}\right.$可求出a与b的值，然后将a、b的值代入二元一次方程组$\left\{\begin{array}{l}{3（x+y）-a（x-y）=16}\\{2（x+y）-b（x-y）=15}\end{array}\right.$即可求出x与y 的值

解答 解：∵$\left\{\begin{array}{l}{3m-an=16}\\{2m-bn=15}\end{array}\right.$的解是$\left\{\begin{array}{l}{m=7}\\{n=1}\end{array}\right.$，
∴$\left\{\begin{array}{l}{21-a=16}\\{14-b=15}\end{array}\right.$，
解得：$\left\{\begin{array}{l}{a=5}\\{b=-1}\end{array}\right.$
把$\left\{\begin{array}{l}{a=5}\\{b=-1}\end{array}\right.$代入$\left\{\begin{array}{l}{3（x+y）-a（x-y）=16}\\{2（x+y）-b（x-y）=15}\end{array}\right.$，
∴化简可得：$\left\{\begin{array}{l}{-2x+8y=16}\\{3x+y=15}\end{array}\right.$
解得：$\left\{\begin{array}{l}{x=4}\\{y=3}\end{array}\right.$，
故答案为：$\left\{\begin{array}{l}{x=4}\\{y=3}\end{array}\right.$

点评 本题考查二元一次方程组的解，解题的关键是熟练运用方程组的解法，本题属于基础题型．