题目内容
4.二元一次方程2x+3y=15有2组正整数解.分析 将x=1,2,…,分别代入2x+3y=15计算得到y为正整数即可.
解答 解:当x=3时,方程变形为6+3y=15,即y=3;
当x=6时,方程变形为12+2y=15,即y=1;
则方程的正整数解有2个.
故答案为:2.
点评 此题考查了解二元一次方程,注意解中x与y必须为正整数.
练习册系列答案
相关题目
16.方程组$\left\{\begin{array}{l}{x+y=1}\\{2y-x=8}\end{array}\right.$的解是( )
| A. | $\left\{\begin{array}{l}{x=-2}\\{y=3}\end{array}\right.$ | B. | $\left\{\begin{array}{l}{x=2}\\{y=-3}\end{array}\right.$ | C. | $\left\{\begin{array}{l}{x=4}\\{y=-3}\end{array}\right.$ | D. | $\left\{\begin{array}{l}{x=-3}\\{y=4}\end{array}\right.$ |