《2025年名校题库八年级数学上册北师大版》

6.(成华区期末)解方程组:
(1)$\left\{\begin{array}{l} 3x+y= 5\textcircled{1},\\ x+3y= 7\textcircled{2};\end{array} \right. $ (2)$\left\{\begin{array}{l} 3(x-1)-y= y+5\textcircled{1},\\ 5(y-1)= 3(x+5)\textcircled{2}.\end{array} \right. $

答案：
(1)解：①×3 - ②，得$9x + 3y - (x + 3y) = 15 - 7$，即$8x = 8$，解得$x = 1$。把$x = 1$代入①，得$3×1 + y = 5$，解得$y = 2$。故原方程组的解为$\left\{\begin{array}{l} x = 1 \\ y = 2 \end{array}\right.$。
(2)解：原方程组整理得$\left\{\begin{array}{l} 3x - 2y = 8① \\ -3x + 5y = 20② \end{array}\right.$。① + ②，得$3y = 28$，解得$y = \frac{28}{3}$。把$y = \frac{28}{3}$代入①，得$3x - 2×\frac{28}{3} = 8$，$3x = 8 + \frac{56}{3} = \frac{24}{3} + \frac{56}{3} = \frac{80}{3}$，解得$x = \frac{80}{9}$。故原方程组的解为$\left\{\begin{array}{l} x = \frac{80}{9} \\ y = \frac{28}{3} \end{array}\right.$。

1.(嘉祥)如果实数x,y满足方程组$\left\{\begin{array}{l} x-2y= -1,\\ x+y= 2,\end{array} \right. 那么(2x-y)^{2022}=$1.

答案： 解：解方程组$\left\{\begin{array}{l} x-2y=-1\\ x+y=2\end{array}\right.$
由$x+y=2$得$x=2-y$，代入$x-2y=-1$，
$2-y-2y=-1$，
$-3y=-3$，
$y=1$。
将$y=1$代入$x=2-y$，得$x=1$。
则$2x-y=2×1 - 1=1$，
所以$(2x-y)^{2022}=1^{2022}=1$。
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2.(武侯区期末)已知关于x,y的二元一次方程组$\left\{\begin{array}{l} mx-3y= 16,\\ 3x-ny= 0\end{array} \right. 的解为\left\{\begin{array}{l} x= 5,\\ y= 3,\end{array} \right. $则关于a,b的二元一次方程组$\left\{\begin{array}{l} m(a+b)-3(a-b)= 16,\\ 3(a+b)-n(a-b)= 0\end{array} \right. $的解是____.$\left\{\begin{array}{l} a=4\\ b=1\end{array}\right.$

答案： 解：将$\left\{\begin{array}{l} x=5\\ y=3\end{array}\right.$代入$\left\{\begin{array}{l} mx - 3y = 16\\ 3x - ny = 0\end{array}\right.$，
得$\left\{\begin{array}{l} 5m - 9 = 16\\ 15 - 3n = 0\end{array}\right.$，解得$\left\{\begin{array}{l} m = 5\\ n = 5\end{array}\right.$。
将$m = 5$，$n = 5$代入关于$a$，$b$的方程组，
得$\left\{\begin{array}{l} 5(a + b)-3(a - b)=16\\ 3(a + b)-5(a - b)=0\end{array}\right.$，
整理得$\left\{\begin{array}{l} 2a + 8b = 16\\ -2a + 8b = 0\end{array}\right.$，
两式相加：$16b = 16$，$b = 1$，
代入$-2a + 8×1 = 0$，得$a = 4$，
所以$\left\{\begin{array}{l} a = 4\\ b = 1\end{array}\right.$。
$\left\{\begin{array}{l} a=4\\ b=1\end{array}\right.$

3.(西川)若方程组$\left\{\begin{array}{l} a_{1}x-b_{1}y= m,\\ a_{2}x-b_{2}y= n\end{array} \right. 的解是\left\{\begin{array}{l} x= 8,\\ y= 10,\end{array} \right. 则方程组\left\{\begin{array}{l} a_{1}(x-2)-b_{1}(y+1)= m,\\ a_{2}(x-2)-b_{2}(y+1)= n\end{array} \right. $的解是$\left\{\begin{array}{l} x=10\\ y=9\end{array}\right.$.

答案： 解：令$x' = x - 2$，$y' = y + 1$，则第二个方程组可化为$\left\{\begin{array}{l} a_{1}x' - b_{1}y' = m\\ a_{2}x' - b_{2}y' = n\end{array}\right.$。
因为已知方程组$\left\{\begin{array}{l} a_{1}x - b_{1}y = m\\ a_{2}x - b_{2}y = n\end{array}\right.$的解是$\left\{\begin{array}{l} x = 8\\ y = 10\end{array}\right.$，所以$\left\{\begin{array}{l} x' = 8\\ y' = 10\end{array}\right.$。
即$\left\{\begin{array}{l} x - 2 = 8\\ y + 1 = 10\end{array}\right.$，解得$\left\{\begin{array}{l} x = 10\\ y = 9\end{array}\right.$。
$\left\{\begin{array}{l} x=10\\ y=9\end{array}\right.$

4.(实外)
(1)解方程组:$\left\{\begin{array}{l} \frac {x+y}{2}+\frac {x-y}{3}= 6,\\ 4(x+y)-5(x-y)= 2.\end{array} \right. $
(2)已知$\left\{\begin{array}{l} a+2b= 4,\\ 3a+2b= 8,\end{array} \right. 求a+b$的值.

答案：
(1)解:$\left\{\begin{array}{l} \frac {x+y}{2}+\frac {x-y}{3}=6①,\\ 4(x+y)-5(x-y)=2②.\end{array}\right.$
由$①×6$，得$3(x+y)+2(x-y)=36③$.
设$m=x+y$，$n=x-y$，则原方程组可化为$\left\{\begin{array}{l} 3m + 2n = 36③\\ 4m - 5n = 2②\end{array}\right.$
由$③×5 + ②×2$，得$15m + 10n + 8m - 10n = 180 + 4$，$23m = 184$，解得$m = 8$。
将$m = 8$代入$②$，得$4×8 - 5n = 2$，$32 - 5n = 2$，解得$n = 6$。
即$\left\{\begin{array}{l} x + y = 8\\ x - y = 6\end{array}\right.$
两式相加，得$2x = 14$，解得$x = 7$。
将$x = 7$代入$x + y = 8$，得$y = 1$。
$\therefore$原方程组的解为$\left\{\begin{array}{l} x = 7\\ y = 1\end{array}\right.$
(2)解:$\left\{\begin{array}{l} a + 2b = 4①\\ 3a + 2b = 8②\end{array}\right.$
由$② - ①$，得$2a = 4$，解得$a = 2$。
将$a = 2$代入$①$，得$2 + 2b = 4$，解得$b = 1$。
$\therefore a + b = 2 + 1 = 3$

1. (石室联中)若方程组$\left\{\begin{array}{l} x+y= 1,\\ 2x+y= 5\end{array} \right. 的解也是方程3x+ky= 0$的解,则$k=$4.

答案： 解：解方程组$\left\{\begin{array}{l} x+y=1\\ 2x+y=5\end{array}\right.$
用第二个方程减第一个方程得：$2x + y - (x + y) = 5 - 1$
$2x + y - x - y = 4$
$x = 4$
把$x = 4$代入$x + y = 1$得：$4 + y = 1$，$y = -3$
所以方程组的解为$\left\{\begin{array}{l} x = 4\\ y = -3\end{array}\right.$
将$\left\{\begin{array}{l} x = 4\\ y = -3\end{array}\right.$代入$3x + ky = 0$得：$3×4 + k×(-3) = 0$
$12 - 3k = 0$
$-3k = -12$
$k = 4$
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2. (高新区期末)若关于$x,y的二元一次方程组\left\{\begin{array}{l} x+y= 5k,\\ x-y= k\end{array} \right. 的解也是二元一次方程2x+3y= 24$的解,则$k$的值为2.

答案： 解：解方程组$\left\{\begin{array}{l} x + y = 5k \\ x - y = k \end{array}\right.$，
两式相加得：$2x = 6k$，解得$x = 3k$，
将$x = 3k$代入$x + y = 5k$，得$3k + y = 5k$，解得$y = 2k$，
把$x = 3k$，$y = 2k$代入$2x + 3y = 24$，得$2×3k + 3×2k = 24$，
即$6k + 6k = 24$，$12k = 24$，解得$k = 2$。
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