Question

19．二元一次方程2x+3y=18的正整数解是$\left\{\begin{array}{l}{x=3}\\{y=4}\end{array}\right.$或$\left\{\begin{array}{l}{x=6}\\{y=2}\end{array}\right.$．

Answer 1

分析 把x看做已知数表示出y，即可确定出方程的正整数解．

解答 解：方程2x+3y=18，
解得：y=-$\frac{2}{3}$x+6，
当x=3时，y=4；x=6时，y=2，
则方程的正整数解为$\left\{\begin{array}{l}{x=3}\\{y=4}\end{array}\right.$或$\left\{\begin{array}{l}{x=6}\\{y=2}\end{array}\right.$，
故答案为：$\left\{\begin{array}{l}{x=3}\\{y=4}\end{array}\right.$或$\left\{\begin{array}{l}{x=6}\\{y=2}\end{array}\right.$

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．