Question

19．方程5x+7y=21有無數(shù)組解；二元一次方程x+3y=7的非負(fù)整數(shù)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 二元一次方程有無數(shù)組解；把y看做已知數(shù)求出x，即可確定出非負(fù)整數(shù)解．

解答 解：方程5x+7y=21，有無數(shù)組解；
方程x+3y=7，解得：x=-3y+7，
當(dāng)y=0時，x=7；y=1時，x=4；y=2時，x=1；
則方程的非負(fù)整數(shù)解有$\left\{\begin{array}{l}{x=7}\\{y=0}\end{array}\right.$；$\left\{\begin{array}{l}{x=4}\\{y=1}\end{array}\right.$；$\left\{\begin{array}{l}{x=1}\\{y=2}\end{array}\right.$，
故答案為：無數(shù)；$\left\{\begin{array}{l}{x=7}\\{y=0}\end{array}\right.$；$\left\{\begin{array}{l}{x=4}\\{y=1}\end{array}\right.$；$\left\{\begin{array}{l}{x=1}\\{y=2}\end{array}\right.$

點(diǎn)評 此題考查了解二元一次方程，熟練掌握運(yùn)算法則是解本題的關(guān)鍵．