《2025年名校题库八年级数学上册北师大版》

7.(嘉祥)
(1)若方程组$\left\{\begin{array}{l} 4x+y= k+1,\\ x+4y= 3\end{array} \right. 的解满足条件0\lt x+y\lt 1$,则$k$的取值范围是多少?
(2)甲、乙两位同学解关于$x,y的方程组\left\{\begin{array}{l} ax+by= 2,\\ cx-7y= 7,\end{array} \right. 甲正确地解得\left\{\begin{array}{l} x= 3,\\ y= 2;\end{array} \right. 乙因把c$看错了,解得$\left\{\begin{array}{l} x= -2,\\ y= 2.\end{array} \right. $根据上述条件,你认为能否求出原方程组中$a,b,c$的值? 若不能,请说明理由.

答案：
(1)解：将方程组中的两个方程相加，得$5x + 5y = k + 4$，则$x + y=\frac{k + 4}{5}$。
因为$0\lt x + y\lt1$，所以$0\lt\frac{k + 4}{5}\lt1$。
不等式两边同时乘以$5$，得$0\lt k + 4\lt5$。
不等式两边同时减去$4$，得$-4\lt k\lt1$。
(2)解：能。
因为甲正确解得$\begin{cases}x = 3\\y = 2\end{cases}$，代入方程组$\begin{cases}ax + by = 2\\cx - 7y = 7\end{cases}$，得$\begin{cases}3a + 2b = 2\\3c - 7×2 = 7\end{cases}$，由$3c - 14 = 7$，解得$c = 7$。
乙因把$c$看错，解得$\begin{cases}x=-2\\y = 2\end{cases}$，但$a$，$b$的值正确，代入$ax + by = 2$，得$-2a + 2b = 2$。
联立$\begin{cases}3a + 2b = 2\\-2a + 2b = 2\end{cases}$，用第一个方程减去第二个方程，得$5a = 0$，解得$a = 0$。
将$a = 0$代入$-2a + 2b = 2$，得$2b = 2$，解得$b = 1$。
综上，$a = 0$，$b = 1$，$c = 7$。

8.(七中育才)已知关于$x,y的方程组\left\{\begin{array}{l} ax+2y= 1+a,\\ 2x+2(a-1)y= 3,\end{array} \right. 分别求出当a$为何值时,方程组:(1)有唯一一组解;(2)无解;(3)有无穷多组解.

答案： 解:$\left\{\begin{array}{l} ax + 2y = 1 + a\enclose{circle}{1},\\ 2x + 2(a - 1)y = 3\enclose{circle}{2}.\end{array}\right.$
由$\enclose{circle}{1}×(a - 1)$,得$a(a - 1)x + 2(a - 1)y = (1 + a)(a - 1)\enclose{circle}{3}$.
由$\enclose{circle}{3} - \enclose{circle}{2}$,得$a(a - 1)x - 2x = a^{2} - 1 - 3$,
$\therefore (a^{2} - a - 2)x = a^{2} - 4$,
$\therefore (a - 2)(a + 1)x = (a + 2)(a - 2)$.
(1)当$(a - 2)(a + 1) ≠ 0$,即$a ≠ 2$且$a ≠ -1$时,方程有唯一解,原方程组有唯一一组解.
(2)当$(a - 2)(a + 1) = 0$且$(a + 2)(a - 2) ≠ 0$时,即$a = -1$时,方程无解,原方程组无解.
(3)当$(a - 2)(a + 1) = 0$且$(a + 2)(a - 2) = 0$时,即$a = 2$时,方程有无穷多解,原方程组有无穷多组解.

1. (高新区期末)方程组$\left\{\begin{array}{l} x+y+z= 23,\\ x-y= 1,\\ 2x+y-z= 20\end{array} \right. $的解是$\left\{\begin{array}{l} x=9,\\ y=8,\\ z=6\end{array}\right.$.

答案： 解：$\left\{\begin{array}{l} x+y+z=23,①\\ x-y=1,②\\ 2x+y-z=20,③\end{array}\right.$
①+③，得$3x+2y=43$，④
由②得$x=y+1$，⑤
把⑤代入④，得$3(y+1)+2y=43$，解得$y=8$，
把$y=8$代入⑤，得$x=9$，
把$x=9$，$y=8$代入①，得$9+8+z=23$，解得$z=6$，
所以方程组的解是$\left\{\begin{array}{l} x=9,\\ y=8,\\ z=6.\end{array}\right.$

2. (嘉祥)已知关于$x,y,z的方程组\left\{\begin{array}{l} 3x-y= 5,\\ 2x+y-z= 0,\\ 4a+5by-z= -22\end{array} \right. 与关于x,y,z的方程组\left\{\begin{array}{l} ax-by+z= 8,\\ x+y+5z= c,\\ 2x+3y= -4\end{array} \right. $有相同的解,则$a= $______,$b= $______,$c= $______.
23-1

答案： 解：联立方程组$\left\{\begin{array}{l}3x - y = 5\\2x + 3y = -4\end{array}\right.$，
由$3x - y = 5$得$y = 3x - 5$，代入$2x + 3y = -4$，
$2x + 3(3x - 5) = -4$，
$2x + 9x - 15 = -4$，
$11x = 11$，$x = 1$，
则$y = 3×1 - 5 = -2$。
将$x = 1$，$y = -2$代入$2x + y - z = 0$，
$2×1 + (-2) - z = 0$，$z = 0$。
把$x = 1$，$y = -2$，$z = 0$代入$\left\{\begin{array}{l}ax - by + z = 8\\4a + 5by - z = -22\\x + y + 5z = c\end{array}\right.$，
得$\left\{\begin{array}{l}a + 2b = 8\\4a - 10b = -22\\c = 1 + (-2) + 0\end{array}\right.$。
解$\left\{\begin{array}{l}a + 2b = 8\\4a - 10b = -22\end{array}\right.$，
由$a = 8 - 2b$代入$4(8 - 2b) - 10b = -22$，
$32 - 8b - 10b = -22$，$-18b = -54$，$b = 3$，
则$a = 8 - 6 = 2$。
$c = -1$。
$a = 2$，$b = 3$，$c = -1$。
答案：2；3；-1