Question

5．对于二元一次方程x+3y=10，请你写出一组正整数解$\left\{\begin{array}{l}{x=7}\\{y=1}\end{array}\right.$或$\left\{\begin{array}{l}{x=4}\\{y=2}\end{array}\right.$或$\left\{\begin{array}{l}{x=1}\\{y=3}\end{array}\right.$．

Answer 1

分析 把y看做已知数求出解表示出x，即可确定出正整数解．

解答 解：方程x+3y=10，
解得：x=-3y+10，
当y=1时，x=7；y=2时，x=4；y=3时，x=1，
则方程的正整数解为$\left\{\begin{array}{l}{x=7}\\{y=1}\end{array}\right.$或$\left\{\begin{array}{l}{x=4}\\{y=2}\end{array}\right.$或$\left\{\begin{array}{l}{x=1}\\{y=3}\end{array}\right.$．
故答案为：$\left\{\begin{array}{l}{x=7}\\{y=1}\end{array}\right.$或$\left\{\begin{array}{l}{x=4}\\{y=2}\end{array}\right.$或$\left\{\begin{array}{l}{x=1}\\{y=3}\end{array}\right.$

点评 此题考查了解二元一次方程，解题的关键是将一个未知数看做已知数求出另一个未知数．