Question

2．方程x+2y=7的正整数解有3组，分别为$\left\{\begin{array}{l}{x=1}\\{y=3}\end{array}\right.$；$\left\{\begin{array}{l}{x=3}\\{y=2}\end{array}\right.$；$\left\{\begin{array}{l}{x=5}\\{y=-1}\end{array}\right.$．

Answer 1

分析 把x看做已知数求出y，确定出正整数解即可．

解答 解：方程x+2y=7，
解得：y=$\frac{7-x}{2}$，
当x=1时，y=3；x=3，y=2；x=5，y=1，
则方程的正整数解有3组，分别为$\left\{\begin{array}{l}{x=1}\\{y=3}\end{array}\right.$；$\left\{\begin{array}{l}{x=3}\\{y=2}\end{array}\right.$；$\left\{\begin{array}{l}{x=5}\\{y=-1}\end{array}\right.$，
故答案为：3；$\left\{\begin{array}{l}{x=1}\\{y=3}\end{array}\right.$；$\left\{\begin{array}{l}{x=3}\\{y=2}\end{array}\right.$；$\left\{\begin{array}{l}{x=5}\\{y=-1}\end{array}\right.$

点评 此题考查了解二元一次方程，解题的关键是将x看做已知数求出y．