《2025年轻松总复习暑假作业七年级数学》

【例题1】求方程$3x+2y= 17$的正整数解.

答案： 解:将方程变形为$y= \frac {17-3x}{2}.\because y＞0,\therefore \frac {17-3x}{2}＞0$,解得$x＜\frac {17}{3}$,即$0＜x＜\frac {17}{3}.\because y$为正整数,即$\frac {17-3x}{2}$为正整数,$\therefore 17-3x$为偶数,$\therefore x$必为奇数,$\therefore x= 1,3,5$.当$x= 1$时,$y= \frac {17-3x}{2}= \frac {17-3×1}{2}= 7$;当$x= 3$时,$y= \frac {17-3x}{2}= \frac {17-3×3}{2}= 4$;当$x= 5$时,$y= \frac {17-3x}{2}= \frac {17-3×5}{2}= 1$.故原方程的正整数解为$\left\{\begin{array}{l} x= 1,\\ y= 7,\end{array} \right. \left\{\begin{array}{l} x= 3,\\ y= 4,\end{array} \right. \left\{\begin{array}{l} x= 5,\\ y= 1.\end{array} \right. $

2. 利用方程组解的情况,求字母的值(取值范围)
【例题2】若关于x,y的二元一次方程组$\left\{\begin{array}{l} x-3y= 4m+3,\\ x+5y= 5\end{array} \right. 的解满足x+y≤0$,则m的取值范围是____.

答案：

解析:$\left\{ \begin{array} { l } { x - 3 y = 4 m + 3 , \textcircled { 1 } } \\ { x + 5 y = 5 . \textcircled { 2 } } \end{array} \right.$ ①+②,得$2x+2y= 4m+8$,则$x+y= 2m+4.\because x+y≤0$,$\therefore 2m+4≤0$,解得$m≤-2$.

答案:$m≤-2$