Question

19．若$\left\{\begin{array}{l}{x=2}\\{y=1}\end{array}\right.$是方程组$\left\{\begin{array}{l}{mx-ny=1}\\{nx+my=8}\end{array}\right.$的解，则m、n的值是$\left\{\begin{array}{l}{m=2}\\{n=3}\end{array}\right.$．

Answer 1

分析 先把$\left\{\begin{array}{l}{x=2}\\{y=1}\end{array}\right.$代入方程，可得$\left\{\begin{array}{l}{2m-n=1}\\{2n+m=8}\end{array}\right.$，解可求m、n的值，最后把m、n的值代入所求代数式计算即可．

解答 解：先把$\left\{\begin{array}{l}{x=2}\\{y=1}\end{array}\right.$代入方程，可得$\left\{\begin{array}{l}{2m-n=1}\\{2n+m=8}\end{array}\right.$，
解得：$\left\{\begin{array}{l}{m=2}\\{n=3}\end{array}\right.$．
故答案为：$\left\{\begin{array}{l}{m=2}\\{n=3}\end{array}\right.$．

点评 本题考查了二元一次方程的解，解题的关键是掌握加减消元的思想．