Question

5．寫出方程x+3y=10正整數(shù)解$\left\{\begin{array}{l}{{x}_{1}=7}\\{{y}_{1}=1}\end{array}\right.$，$\left\{\begin{array}{l}{{x}_{2}=4}\\{{y}_{2}=2}\end{array}\right.$，$\left\{\begin{array}{l}{{x}_{3}=1}\\{{y}_{3}=3}\end{array}\right.$．

Answer 1

分析 先變形得出x=10-3y，再取正整數(shù)解即可．

解答 解：x+3y=10，
x=10-3y，
當y=1時，x=7；
當y=2時，x=4，
當y=3時，x=1；
故答案為：$\left\{\begin{array}{l}{{x}_{1}=7}\\{{y}_{1}=1}\end{array}\right.$，$\left\{\begin{array}{l}{{x}_{2}=4}\\{{y}_{2}=2}\end{array}\right.$，$\left\{\begin{array}{l}{{x}_{3}=1}\\{{y}_{3}=3}\end{array}\right.$．

點評 本題考查了解二元一次方程，能求出符合的所有正整數(shù)解是解此題的關鍵．