Question

2．如果x、y都是正整数，则满足x+2y=11的x、y值有5组，分别是$\left\{\begin{array}{l}{x=1}\\{y=5}\end{array}\right.$，$\left\{\begin{array}{l}{x=3}\\{y=4}\end{array}\right.$，$\left\{\begin{array}{l}{x=5}\\{y=3}\end{array}\right.$，$\left\{\begin{array}{l}{x=7}\\{y=2}\end{array}\right.$，$\left\{\begin{array}{l}{x=9}\\{y=1}\end{array}\right.$．

Answer 1

分析 采用列举法求得方程组的解即可．

解答 解：∵①当y=1时，x=9，
②当y=2时，x=7，
③当y=3时，x=5，
④当y=4时，x=3，
⑤当y=5时，x=1，
∴方程x+2y=11的正整数解有5组．
故答案为：5，$\left\{\begin{array}{l}{x=1}\\{y=5}\end{array}\right.$，$\left\{\begin{array}{l}{x=3}\\{y=4}\end{array}\right.$，$\left\{\begin{array}{l}{x=5}\\{y=3}\end{array}\right.$，$\left\{\begin{array}{l}{x=7}\\{y=2}\end{array}\right.$，$\left\{\begin{array}{l}{x=9}\\{y=1}\end{array}\right.$

点评 本题主要考查的是二元一次方程的解，列举法的应用是解题的关键．