Question

19．方程2x+y=7的正整数解是$\left\{\begin{array}{l}{x=1}\\{y=5}\end{array}\right.$，$\left\{\begin{array}{l}{x=2}\\{y=3}\end{array}\right.$，$\left\{\begin{array}{l}{x=3}\\{y=1}\end{array}\right.$．

Answer 1

分析 把x看做已知数表示出y，即可确定出正整数解．

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

点评 此题考查了解二元一次方程，解题的关键是将x看作已知数求出y．