《2025年暑假乐园海南出版社七年级数学人教版》

2. 已知$\left\{\begin{array}{l} ax+by=5,\\ bx+ay=2\end{array}\right. $的解是$\left\{\begin{array}{l} x=4,\\ y=3\end{array}\right. $，则( )。
A. $\left\{\begin{array}{l} a=2,\\ b=1\end{array}\right. $
B. $\left\{\begin{array}{l} a=2,\\ b=-1\end{array}\right. $
C. $\left\{\begin{array}{l} a=-2,\\ b=1\end{array}\right. $
D. $\left\{\begin{array}{l} a=-2,\\ b=-1\end{array}\right. $

答案： B

1. 解下列方程组。
(1)$\left\{\begin{array}{l} x+y=3,\\ 2x-y=6;\end{array}\right. $
(2)$\left\{\begin{array}{l} x-4y=8,\\ 3x+2y=17.\end{array}\right. $

答案： 【解析】：
(1) 对于方程组$\left\{\begin{array}{l}x + y = 3&①\\2x - y = 6&②\end{array}\right.$
将①$+$②可得：
$(x + y)+(2x - y)=3 + 6$
去括号得$x + y+2x - y=3 + 6$
合并同类项得$3x=9$
解得$x = 3$
把$x = 3$代入①式得$3 + y=3$
解得$y=0$
(2) 对于方程组$\left\{\begin{array}{l}x - 4y = 8&①\\3x + 2y = 17&②\end{array}\right.$
由②$\times2$得$6x + 4y = 34$ ③
①$+$③可得：
$(x - 4y)+(6x + 4y)=8 + 34$
去括号得$x - 4y+6x + 4y=8 + 34$
合并同类项得$7x=42$
解得$x = 6$
把$x = 6$代入①式得$6-4y = 8$
移项得$-4y=8 - 6$
即$-4y = 2$
解得$y=-\frac{1}{2}$
【答案】：
(1)$\left\{\begin{array}{l}x = 3\\y = 0\end{array}\right.$；
(2)$\left\{\begin{array}{l}x = 6\\y=-\frac{1}{2}\end{array}\right.$