Question

19．方程5x+7y=21有无数组解；二元一次方程x+3y=7的非负整数解是$\left\{\begin{array}{l}{x=7}\\{y=0}\end{array}\right.$；$\left\{\begin{array}{l}{x=4}\\{y=1}\end{array}\right.$；$\left\{\begin{array}{l}{x=1}\\{y=2}\end{array}\right.$．

Answer 1

分析 二元一次方程有无数组解；把y看做已知数求出x，即可确定出非负整数解．

解答 解：方程5x+7y=21，有无数组解；
方程x+3y=7，解得：x=-3y+7，
当y=0时，x=7；y=1时，x=4；y=2时，x=1；
则方程的非负整数解有$\left\{\begin{array}{l}{x=7}\\{y=0}\end{array}\right.$；$\left\{\begin{array}{l}{x=4}\\{y=1}\end{array}\right.$；$\left\{\begin{array}{l}{x=1}\\{y=2}\end{array}\right.$，
故答案为：无数；$\left\{\begin{array}{l}{x=7}\\{y=0}\end{array}\right.$；$\left\{\begin{array}{l}{x=4}\\{y=1}\end{array}\right.$；$\left\{\begin{array}{l}{x=1}\\{y=2}\end{array}\right.$

点评 此题考查了解二元一次方程，熟练掌握运算法则是解本题的关键．