题目内容
10.关于x,y的方程2x=8-y的正整数解的组数为$\left\{\begin{array}{l}{x=3}\\{y=2}\end{array}\right.$,$\left\{\begin{array}{l}{x=2}\\{y=4}\end{array}\right.$,$\left\{\begin{array}{l}{x=1}\\{y=6}\end{array}\right.$.分析 用y表示出x,即可确定出方程的正整数解.
解答 解:方程整理得:x=$\frac{8-y}{2}$=4-$\frac{y}{2}$,
当y=2时,x=4-1=3;当y=4时,x=4-2=2;当y=6时,x=4-3=1,
则方程的正整数解为$\left\{\begin{array}{l}{x=3}\\{y=2}\end{array}\right.$,$\left\{\begin{array}{l}{x=2}\\{y=4}\end{array}\right.$,$\left\{\begin{array}{l}{x=1}\\{y=6}\end{array}\right.$,
故答案为:$\left\{\begin{array}{l}{x=3}\\{y=2}\end{array}\right.$,$\left\{\begin{array}{l}{x=2}\\{y=4}\end{array}\right.$,$\left\{\begin{array}{l}{x=1}\\{y=6}\end{array}\right.$
点评 此题考查了二元一次方程的解,方程的解即为能使方程左右两边相等的未知数的值.
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