题目内容

7.二元一次方程x+y=6的正整数解为$\left\{\begin{array}{l}{{x}_{1}=1}\\{{y}_{1}=5}\end{array}\right.$,$\left\{\begin{array}{l}{{x}_{2}=2}\\{{y}_{2}=4}\end{array}\right.$,$\left\{\begin{array}{l}{{x}_{3}=3}\\{{y}_{3}=3}\end{array}\right.$,$\left\{\begin{array}{l}{{x}_{4}=4}\\{{y}_{4}=2}\end{array}\right.$,$\left\{\begin{array}{l}{{x}_{5}=5}\\{{y}_{5}=1}\end{array}\right.$.

分析 根据二元一次方程的解的定义,可得出5组一元一次方程x+y=6的正整数解.

解答 解:当x=1时,y=5;
当x=2时,y=4;
当x=3时,y=3;
当x=4时,y=2;
当x=5时,y=1;
∴方程x+y=6的正整数解为:$\left\{\begin{array}{l}{{x}_{1}=1}\\{{y}_{1}=5}\end{array}\right.$,$\left\{\begin{array}{l}{{x}_{2}=2}\\{{y}_{2}=4}\end{array}\right.$,$\left\{\begin{array}{l}{{x}_{3}=3}\\{{y}_{3}=3}\end{array}\right.$,$\left\{\begin{array}{l}{{x}_{4}=4}\\{{y}_{4}=2}\end{array}\right.$,$\left\{\begin{array}{l}{{x}_{5}=5}\\{{y}_{5}=1}\end{array}\right.$;
故答案为:$\left\{\begin{array}{l}{{x}_{1}=1}\\{{y}_{1}=5}\end{array}\right.$,$\left\{\begin{array}{l}{{x}_{2}=2}\\{{y}_{2}=4}\end{array}\right.$,$\left\{\begin{array}{l}{{x}_{3}=3}\\{{y}_{3}=3}\end{array}\right.$,$\left\{\begin{array}{l}{{x}_{4}=4}\\{{y}_{4}=2}\end{array}\right.$,$\left\{\begin{array}{l}{{x}_{5}=5}\\{{y}_{5}=1}\end{array}\right.$.

点评 本题考查了二元一次方程的整数解的情况,在求一个二元一次方程的整数解时,往往采用“给一个,求一个”的方法,即先给出其中一个未知数(一般是系数绝对值较大的)的值,再依次求出另一个的对应值.

练习册系列答案
相关题目

违法和不良信息举报电话:027-86699610 举报邮箱:58377363@163.com

精英家教网