《2025年暑假零距离七年级数学人教版》

13. 在一家水果店,小明买了1斤苹果、4斤西瓜、2斤橙子,共付27.2元;小惠买了2斤苹果、6斤西瓜、2斤橙子,共付32.4元.则买1斤西瓜和1斤橙子需付11元.

答案： 11

14. 某中学现有学生4 200人,计划一年后初中在校学生增加8%,高中在校学生增加11%.这样会使在校学生共增加10%,这所中学现在的初中、高中在校学生分别有多少人?
(1)设现在的初中在校学生有x人,高中在校学生有y人,填写下表:
| |初中在校学生|高中在校学生|总人数|
|现有学生数/人|x|y|4 200|
|一年后学生数/人|(1 + 8%)x|(1 + 11%)y|4200×(1 + 10%)|
(2)列出关于x,y的二元一次方程组:$\begin{cases}x + y = 4200, \\ (1 + 8\%)x + (1 + 11\%)y = 4200×(1 + 10\%)\end{cases}$.

答案：
(1) $(1 + 8\%)x$ $(1 + 11\%)y$ $4200×(1 + 10\%)$
(2) $\begin{cases}x + y = 4200, \\ (1 + 8\%)x + (1 + 11\%)y = 4200×(1 + 10\%)\end{cases}$

15. (6分)解方程组:
(1)$\left\{\begin{array}{l} 3y-4x= 0,\\ 4x+y= 8;\end{array} \right. $
(2)$\left\{\begin{array}{l} 2x+y= 3,\\ \frac {1}{2}x-\frac {3}{2}y= -1;\end{array} \right. $
(3)$\left\{\begin{array}{l} x+2y= 4,\\ 2x+5y-2z= 11,\\ 3x-5y+2z= -1.\end{array} \right. $

答案：
(1)解：$\left\{\begin{array}{l}3y - 4x = 0①\\4x + y = 8②\end{array}\right.$
由①得$4x = 3y$③
把③代入②得$3y + y = 8$
$4y = 8$
$y = 2$
把$y = 2$代入③得$4x = 3×2$
$x=\frac{3}{2}$
$\therefore\left\{\begin{array}{l}x = \frac{3}{2}\\y = 2\end{array}\right.$
(2)解：$\left\{\begin{array}{l}2x + y = 3①\\frac{1}{2}x-\frac{3}{2}y=-1②\end{array}\right.$
②×2得$x - 3y=-2$③
①×3得$6x + 3y = 9$④
③+④得$7x = 7$
$x = 1$
把$x = 1$代入①得$2×1 + y = 3$
$y = 1$
$\therefore\left\{\begin{array}{l}x = 1\\y = 1\end{array}\right.$
(3)解：$\left\{\begin{array}{l}x + 2y = 4①\\2x + 5y - 2z = 11②\\3x - 5y + 2z=-1③\end{array}\right.$
②+③得$5x = 10$
$x = 2$
把$x = 2$代入①得$2 + 2y = 4$
$2y = 2$
$y = 1$
把$x = 2,y = 1$代入②得$2×2 + 5×1-2z = 11$
$4 + 5-2z = 11$
$-2z = 2$
$z=-1$
$\therefore\left\{\begin{array}{l}x = 2\\y = 1\\z=-1\end{array}\right.$

16. (6分)在解方程组$\left\{\begin{array}{l} 2x+5y= 3,\enclose{circle} {1}\\ 4x+11y= 5\enclose{circle} {2}\end{array} \right. $时,小明采用了一种“整体代换”的解法.
解:将方程②变形,得$4x+10y+y= 5,即2(2x+5y)+y= 5$,③
把方程①代入③,得$2×3+y= 5,$
$\therefore y= -1.$
把$y= -1$代入①,得$2x+5×(-1)= 3$,解得$x= 4,\therefore 方程组的解为\left\{\begin{array}{l} x= 4,\\ y= -1.\end{array} \right. $
请你解决以下问题:模仿小明的“整体代换”法解方程组$\left\{\begin{array}{l} 4x-3y= 6,\\ 8x-7y= 18.\end{array} \right. $

答案： 解：$\begin{cases}4x - 3y = 6, ① \\ 8x - 7y = 18, ②\end{cases}$
将②变形，得$2(4x - 3y) - y = 18$，③
把①代入③，得$2×6 - y = 18$，
$\therefore y = -6$。
把$y = -6$代入①，得$4x - 3×(-6) = 6$，解得$x = -3$，
$\therefore$ 方程组的解为$\begin{cases}x = -3, \\ y = -6\end{cases}$

17. (8分)(1)若在方程$2x-y= \frac {1}{3}$的解中,x,y互为相反数,求xy的值;
(2)已知$\left\{\begin{array}{l} x= 2,\\ y= 1\end{array} \right. 是方程组\left\{\begin{array}{l} 2x+(m-1)y= 2,\\ nx+y= 1\end{array} \right. $的解,求$m+n$的值.

答案： 解：
(1) $\because x$，$y$ 互为相反数，$\therefore x + y = 0$，$\therefore \begin{cases}2x - y = \frac{1}{3}, \\ x + y = 0\end{cases}$ 解得 $\begin{cases}x = \frac{1}{9}, \\ y = -\frac{1}{9}\end{cases}$ $\therefore xy = \frac{1}{9}×(-\frac{1}{9}) = -\frac{1}{81}$
(2) 把 $\begin{cases}x = 2, \\ y = 1\end{cases}$ 代入方程组 $\begin{cases}2x + (m - 1)y = 2, \\ nx + y = 1\end{cases}$ 得 $\begin{cases}4 + m - 1 = 2, \\ 2n + 1 = 1\end{cases}$ 解得 $\begin{cases}m = -1, \\ n = 0\end{cases}$ $\therefore m + n = -1 + 0 = -1$