Question

5．對于二元一次方程x+3y=10，請你寫出一組正整數(shù)解$\left\{\begin{array}{l}{x=7}\\{y=1}\end{array}\right.$或$\left\{\begin{array}{l}{x=4}\\{y=2}\end{array}\right.$或$\left\{\begin{array}{l}{x=1}\\{y=3}\end{array}\right.$．

Answer 1

分析 把y看做已知數(shù)求出解表示出x，即可確定出正整數(shù)解．

解答 解：方程x+3y=10，
解得：x=-3y+10，
當y=1時，x=7；y=2時，x=4；y=3時，x=1，
則方程的正整數(shù)解為$\left\{\begin{array}{l}{x=7}\\{y=1}\end{array}\right.$或$\left\{\begin{array}{l}{x=4}\\{y=2}\end{array}\right.$或$\left\{\begin{array}{l}{x=1}\\{y=3}\end{array}\right.$．
故答案為：$\left\{\begin{array}{l}{x=7}\\{y=1}\end{array}\right.$或$\left\{\begin{array}{l}{x=4}\\{y=2}\end{array}\right.$或$\left\{\begin{array}{l}{x=1}\\{y=3}\end{array}\right.$

點評 此題考查了解二元一次方程，解題的關鍵是將一個未知數(shù)看做已知數(shù)求出另一個未知數(shù)．