《2025年综合应用创新题典中点八年级数学上册北师大版》

1. 解下列方程组：
(1)$\left\{\begin{array}{l} \frac {x+1}{3}= 2y,\\ 2(x+1)-y= 11;\end{array}\right. $
【解】原方程组整理，得$\left\{\begin{array}{l} x=6y - 1,\enclose{circle} {1}\\ 2x - y = 9,\enclose{circle} {2}\end{array}\right.$
把①代入②，得$2(6y - 1) - y = 9$，解得$y = $1，
把$y = 1$代入①，得$x = $5，
所以原方程组的解为$\left\{\begin{array}{l} x = 5,\\ y = 1.\end{array}\right.$
(2)$\left\{\begin{array}{l} 2x+3y= 7,\\ x-2y= 3.\end{array}\right. $
【解】$\left\{\begin{array}{l} 2x + 3y = 7,\enclose{circle} {1}\\ x - 2y = 3,\enclose{circle} {2}\end{array}\right.$
① - ②×2，得$7y = 1$，解得$y = $$\frac{1}{7}$，
将$y = \frac{1}{7}$代入②，得$x - \frac{2}{7} = 3$，解得$x = $$\frac{23}{7}$，
所以原方程组的解为$\left\{\begin{array}{l} x = \frac{23}{7},\\ y = \frac{1}{7}.\end{array}\right.$

答案： 【解】
(1)原方程组整理，得$\left\{\begin{array}{l} x=6y - 1,\enclose{circle} {1}\\ 2x - y = 9,\enclose{circle} {2}\end{array}\right.$
把①代入②，得$2(6y - 1) - y = 9$，解得$y = 1$，
把$y = 1$代入①，得$x = 5$，
所以原方程组的解为$\left\{\begin{array}{l} x = 5,\\ y = 1.\end{array}\right.$
(2)$\left\{\begin{array}{l} 2x + 3y = 7,\enclose{circle} {1}\\ x - 2y = 3,\enclose{circle} {2}\end{array}\right.$
① - ②×2，得$7y = 1$，解得$y = \frac{1}{7}$，
将$y = \frac{1}{7}$代入②，得$x - \frac{2}{7} = 3$，解得$x = \frac{23}{7}$，
所以原方程组的解为$\left\{\begin{array}{l} x = \frac{23}{7},\\ y = \frac{1}{7}.\end{array}\right.$

2. 解方程组：$\left\{\begin{array}{l} 7x-8y= 22,\\ 3x-5y= 11.\end{array}\right. $
解：
$\begin{cases}7x - 8y = 22&(1)\\3x - 5y = 11&(2)\end{cases}$
$(1)×3$得：$21x - 24y = 66$ $(3)$
$(2)×7$得：$21x - 35y = 77$ $(4)$
$(3) - (4)$得：
$\begin{aligned}(21x - 24y) - (21x - 35y)&=66 - 77\\21x - 24y - 21x + 35y&=-11\\11y&=-11\\y&=-1\end{aligned}$
把$y = -1$代入$(2)$得：
$\begin{aligned}3x - 5×(-1)&=11\\3x + 5&=11\\3x&=11 - 5\\3x&=6\\x&=2\end{aligned}$
所以方程组的解为$\begin{cases} x = 2 \\ y = -1 \end{cases}$。

答案： 解：
$\begin{cases}7x - 8y = 22&(1)\\3x - 5y = 11&(2)\end{cases}$
$(1)×3$得：$21x - 24y = 66$ $(3)$
$(2)×7$得：$21x - 35y = 77$ $(4)$
$(3) - (4)$得：
$\begin{aligned}(21x - 24y) - (21x - 35y)&=66 - 77\\21x - 24y - 21x + 35y&=-11\\11y&=-11\\y&=-1\end{aligned}$
把$y = -1$代入$(2)$得：
$\begin{aligned}3x - 5×(-1)&=11\\3x + 5&=11\\3x&=11 - 5\\3x&=6\\x&=2\end{aligned}$
所以方程组的解为$\begin{cases} x = 2 \\ y = -1 \end{cases}$。

3. 新考法 图表信息法 已知关于x,y的二元一次方程$ax+by= c$的解如下表：
| x | … | -4 | -3 | -2 | -1 | 0 | 1 | … |
| y | … | $\frac {14}{3}$ | 4 | $\frac {10}{3}$ | $\frac {8}{3}$ | 2 | $\frac {4}{3}$ | … |
关于x,y的二元一次方程$mx-ny= k$的解如下表：
| x | … | -4 | -3 | -2 | -1 | 0 | 1 | … |
| y | … | $\frac {11}{2}$ | 4 | $\frac {5}{2}$ | 1 | $-\frac {1}{2}$ | -2 | … |
则关于x,y的二元一次方程组$\left\{\begin{array}{l} a(x+y)+b(x-y)= c,\\ m(x+y)-n(x-y)= k\end{array}\right. $的解为$\left\{\begin{array}{l} x = \frac{1}{2},\\ y = - \frac{7}{2}\end{array}\right.$.

答案： $\left\{\begin{array}{l} x = \frac{1}{2},\\ y = \frac{7}{2}\end{array}\right.$【点拨】由题表可知方程组$\left\{\begin{array}{l} ax + by = c,\\ mx - ny = k\end{array}\right.$的解为$\left\{\begin{array}{l} x = - 3,\\ y = 4,\end{array}\right.$所以由二元一次方程组$\left\{\begin{array}{l} a(x + y) + b(x - y) = c,\\ m(x + y) - n(x - y) = k\end{array}\right.$可得$\left\{\begin{array}{l} x + y = - 3,\\ x - y = 4,\end{array}\right.$解得$\left\{\begin{array}{l} x = \frac{1}{2},\\ y = - \frac{7}{2}.\end{array}\right.$所以方程组$\left\{\begin{array}{l} a(x + y) + b(x - y) = c,\\ m(x + y) - n(x - y) = k\end{array}\right.$的解为$\left\{\begin{array}{l} x = \frac{1}{2},\\ y = - \frac{7}{2}.\end{array}\right.$

4. 新考法 阅读类比法 先阅读,然后解方程组.
解方程组$\begin{cases}{ x-y-1= 0,①}\\{4(x-y)-y= 5，②}\end{cases}$时,可由①得$x - y = 1$,③
再将③代入②,得$4×1 - y = 5$,解得$y = -1$,
从而进一步求得$\begin{cases} x=0 \\ y=-1 \end{cases}$这种方法被称为“整体代入法”.
请用这样的方法解方程组$\begin{cases} 2x - 3y - 2 = 0, \\frac{2x - 3y + 5}{7} + 2y = 9. \end{cases}$
解：由$2x - 3y - 2 = 0$得$2x - 3y = 2$③
将③代入$\frac{2x - 3y + 5}{7}+2y = 9$，得$\frac{2 + 5}{7}+2y = 9$
即$1 + 2y = 9$，
移项可得$2y = 9 - 1$，
即$2y = 8$，
解得$y = 4$。
把$y = 4$代入③得$2x - 3×4 = 2$，
即$2x - 12 = 2$，
移项可得$2x = 2 + 12$，
即$2x = 14$，
解得$x = 7$。
所以方程组的解为$\begin{cases}x = 7\\y = 4\end{cases}$。

答案： 解：由$2x - 3y - 2 = 0$得$2x - 3y = 2$ ③
将③代入$\frac{2x - 3y + 5}{7}+2y = 9$，得$\frac{2 + 5}{7}+2y = 9$
即$1 + 2y = 9$，
移项可得$2y = 9 - 1$，
即$2y = 8$，
解得$y = 4$。
把$y = 4$代入③得$2x - 3×4 = 2$，
即$2x - 12 = 2$，
移项可得$2x = 2 + 12$，
即$2x = 14$，
解得$x = 7$。
所以方程组的解为$\begin{cases}x = 7\\y = 4\end{cases}$。

5. 三名同学对问题“若方程组$\left\{\begin{array}{l} a_{1}x+b_{1}y= c_{1},\\ a_{2}x+b_{2}y= c_{2}\end{array}\right. 的解是\left\{\begin{array}{l} x= 3,\\ y= 4,\end{array}\right. 求方程组\left\{\begin{array}{l} 3a_{1}x+2b_{1}y= 5c_{1},\\ 3a_{2}x+2b_{2}y= 5c_{2}\end{array}\right. $的解”提出了各自的想法. 甲说：“这个题目好像条件不够,不能求解.”乙说：“它们的系数有一定的规律,可以试试.”丙说：“能不能把第二个方程组的两个方程的两边都除以5,通过换元的方法来解决.”参考他们的讨论,谈谈你的看法(若不能求解,请说明原因；若能够求解,请写出求解过程).
【解】可以求解。
因为$\left\{\begin{array}{l} 3a_{1}x + 2b_{1}y = 5c_{1},\\ 3a_{2}x + 2b_{2}y = 5c_{2}\end{array}\right.$可变形为$\left\{\begin{array}{l} \frac{3}{5}a_{1}x + \frac{2}{5}b_{1}y = c_{1},\\ \frac{3}{5}a_{2}x + \frac{2}{5}b_{2}y = c_{2},\end{array}\right.$①
设$m = \frac{3}{5}x$，$n = \frac{2}{5}y$，
所以方程组①可变为$\left\{\begin{array}{l} a_{1}m + b_{1}n = c_{1},\\ a_{2}m + b_{2}n = c_{2}.\end{array}\right.$②
又因为$\left\{\begin{array}{l} a_{1}x + b_{1}y = c_{1},\\ a_{2}x + b_{2}y = c_{2}\end{array}\right.$的解是$\left\{\begin{array}{l} x = 3,\\ y = 4,\end{array}\right.$
所以方程组②的解是$\left\{\begin{array}{l} m = 3,\\ n = 4.\end{array}\right.$
所以$\frac{3}{5}x = 3$，$\frac{2}{5}y = 4$。所以$x = 5$，$y = 10$。
所以方程组$\left\{\begin{array}{l} 3a_{1}x + 2b_{1}y = 5c_{1},\\ 3a_{2}x + 2b_{2}y = 5c_{2}\end{array}\right.$的解是$\left\{\begin{array}{l} x = 5,\\ y = 10.\end{array}\right.$

答案： 【解】可以求解。
因为$\left\{\begin{array}{l} 3a_{1}x + 2b_{1}y = 5c_{1},\\ 3a_{2}x + 2b_{2}y = 5c_{2}\end{array}\right.$可变形为$\left\{\begin{array}{l} \frac{3}{5}a_{1}x + \frac{2}{5}b_{1}y = c_{1},\\ \frac{3}{5}a_{2}x + \frac{2}{5}b_{2}y = c_{2},\end{array}\right.$①
设$m = \frac{3}{5}x$，$n = \frac{2}{5}y$，
所以方程组①可变为$\left\{\begin{array}{l} a_{1}m + b_{1}n = c_{1},\\ a_{2}m + b_{2}n = c_{2}.\end{array}\right.$②
又因为$\left\{\begin{array}{l} a_{1}x + b_{1}y = c_{1},\\ a_{2}x + b_{2}y = c_{2}\end{array}\right.$的解是$\left\{\begin{array}{l} x = 3,\\ y = 4,\end{array}\right.$
所以方程组②的解是$\left\{\begin{array}{l} m = 3,\\ n = 4.\end{array}\right.$
所以$\frac{3}{5}x = 3$，$\frac{2}{5}y = 4$。所以$x = 5$，$y = 10$。
所以方程组$\left\{\begin{array}{l} 3a_{1}x + 2b_{1}y = 5c_{1},\\ 3a_{2}x + 2b_{2}y = 5c_{2}\end{array}\right.$的解是$\left\{\begin{array}{l} x = 5,\\ y = 10.\end{array}\right.$