《2025年暑假衔接七年级数学人教版延边人民出版社》

16. 若关于$x,y$的二元一次方程组$\left\{\begin{array}{l} 3x-my=5,\\ 2x+ny=6\end{array}\right. $的解是$\left\{\begin{array}{l} x=1,\\ y=2,\end{array}\right. $则关于$a,b$的二元一次方程组$\left\{\begin{array}{l} 3(a+b)-m(a-b)=5,\\ 2(a+b)+n(a-b)=6\end{array}\right. $的解是$\left\{\begin{array}{l} a = \frac{3}{2},\\ b = -\frac{1}{2}\end{array}\right.$.

答案： $\left\{\begin{array}{l} a = \frac{3}{2},\\ b = -\frac{1}{2}\end{array}\right.$

17. 有甲、乙、丙三件商品，若购买甲商品 3 件、乙商品 2 件、丙商品 1 件，则共需 315 元；若购买甲商品 1 件、乙商品 2 件、丙商品 3 件，则共需 285 元. 购买甲、乙、丙商品各 1 件共需150元.

答案： 150

18. 解下列方程组：
（1）$\left\{\begin{array}{l} 5x+2y=11,\\ 4x-4y=6;\end{array}\right. $ $\left\{\begin{array}{l} x = 2,\\ y = \frac{1}{2}\end{array}\right.$
（2）$\left\{\begin{array}{l} 3x-2(x+2y)=3,\\ 11x+4(x+2y)=45.\end{array}\right. $ $\left\{\begin{array}{l} x = 3,\\ y = 0\end{array}\right.$

答案：
(1)$\left\{\begin{array}{l} x = 2,\\ y = \frac{1}{2}\end{array}\right.$
(2)$\left\{\begin{array}{l} x = 3,\\ y = 0\end{array}\right.$

19. 已知关于$x,y$的方程组$\left\{\begin{array}{l} 3x-y=5,\\ 4ax+5by=-22\end{array}\right. $与$\left\{\begin{array}{l} 2x-3y+4=0,\\ ax-by-8=0\end{array}\right. $有相同的解，求$a,b$的值.
$a=$$\frac{14}{19}$，$b=$$-\frac{21}{11}$.

答案：
∵两个方程组有相同的解，
∴原方程组可化为①$\left\{\begin{array}{l} 3x - y = 5,\\ 2x - 3y + 4 = 0\end{array}\right.$和②$\left\{\begin{array}{l} 4ax + 5by = -22,\\ ax - by - 8 = 0,\end{array}\right.$解方程组①得$\left\{\begin{array}{l} x = \frac{19}{7},\\ y = \frac{22}{7},\end{array}\right.$代入②得$\left\{\begin{array}{l} a = \frac{14}{19},\\ b = -\frac{21}{11}.\end{array}\right.$

20. 甲、乙两人共同解方程组$\left\{\begin{array}{l} ax+5y=15,①\\ 4x=by-2②\end{array}\right. $时，甲看错了方程①中的$a$，解得$\left\{\begin{array}{l} x=-3,\\ y=-1,\end{array}\right. $乙看错了方程②中的$b$，解得$\left\{\begin{array}{l} x=5,\\ y=4.\end{array}\right. $求$a^{2022}+(-\frac {b}{10})^{2023}$的值.
0

答案： 把$\left\{\begin{array}{l} x = -3,\\ y = -1\end{array}\right.$代入方程②，得$4×(-3) = b·(-1) - 2$，解得$b = 10$。把$\left\{\begin{array}{l} x = 5,\\ y = 4\end{array}\right.$代入方程①，得$5a + 5×4 = 15$，解得$a = -1$，所以$a^{2022} + (-\frac{b}{10})^{2023} = 0$。