题目内容

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

分析 将x看做已知数求出y,即可确定出正整数解.

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

点评 此题考查了解二元一次方程,熟练掌握运算法则是解本题的关键.

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