Question

13．二元一次方程3x+y=7的正整数解是$\left\{\begin{array}{l}x=1\\ y=4\end{array}\right.$$和\left\{\begin{array}{l}x=2\\ y=1\end{array}\right.$．

Answer 1

分析 采用列举法求得方程的所有正整数解即可．

解答 解：∵当x=1时，3+y=7，则y=4；当x=2式，6+y=7，则y=1，
∴方程3x+y=7的正整数解$\left\{\begin{array}{l}{x=1}\\{y=4}\end{array}\right.$和$\left\{\begin{array}{l}{x=2}\\{y=1}\end{array}\right.$．
故答案为：$\left\{\begin{array}{l}{x=1}\\{y=4}\end{array}\right.$和$\left\{\begin{array}{l}{x=2}\\{y=1}\end{array}\right.$．

点评 本题主要考查的是二元一次方程的解，应用列举法求解是解题的关键．