7. 若方程组$\{\begin{array}{l}x-(c+3)xy=3,\\ x^{a - 2}-y^{b + 3}=4\end{array} $是关于$x,y$的二元一次方程组，则代数式$a + b + c$的值是 ______ .
答案：7.$-2$或$-3$ 解析：由二元一次方程组的概念，得$c + 3 = 0$，$a - 2 = 1$，$b + 3 = 1$，解得$c = - 3$，$a = 3$，$b = - 2$，所以$a + b + c = - 2$。或$c + 3 = 0$，$a - 2 = 0$，$b + 3 = 1$，解得$c = - 3$，$a = 2$，$b = - 2$，所以$a + b + c = - 3$。
解析：
因为方程组$\{\begin{array}{l}x-(c+3)xy=3\\x^{a - 2}-y^{b + 3}=4\end{array} $是关于$x,y$的二元一次方程组，所以：
情况一：$c + 3 = 0$，$a - 2 = 1$，$b + 3 = 1$
解得$c=-3$，$a=3$，$b=-2$，则$a + b + c=3+(-2)+(-3)=-2$
情况二：$c + 3 = 0$，$a - 2 = 0$，$b + 3 = 1$
解得$c=-3$，$a=2$，$b=-2$，则$a + b + c=2+(-2)+(-3)=-3$
$-2$或$-3$
情况一：$c + 3 = 0$，$a - 2 = 1$，$b + 3 = 1$
解得$c=-3$，$a=3$，$b=-2$，则$a + b + c=3+(-2)+(-3)=-2$
情况二：$c + 3 = 0$，$a - 2 = 0$，$b + 3 = 1$
解得$c=-3$，$a=2$，$b=-2$，则$a + b + c=2+(-2)+(-3)=-3$
$-2$或$-3$
8. 已知方程组$\{\begin{array}{l}ax+by=2,\\ cx+dy=3\end{array} $的解是$\{\begin{array}{l}x=2,\\ y=3\end{array} $则方程组$\{\begin{array}{l}a(x - 1)+b(y - 1)=2,\\ c(x - 1)+d(y - 1)=3\end{array} $的解为 ______ .
答案：8.$\begin{cases}x = 3,\\y = 4\end{cases}$
解析：
令$m = x - 1$，$n = y - 1$，则方程组$\{\begin{array}{l}a(x - 1)+b(y - 1)=2,\\ c(x - 1)+d(y - 1)=3\end{array}$可化为$\{\begin{array}{l}am + bn = 2,\\ cm + dn = 3\end{array}$。
已知原方程组$\{\begin{array}{l}ax + by = 2,\\ cx + dy = 3\end{array}$的解是$\{\begin{array}{l}x = 2,\\ y = 3\end{array}$，所以新方程组$\{\begin{array}{l}am + bn = 2,\\ cm + dn = 3\end{array}$的解为$\{\begin{array}{l}m = 2,\\ n = 3\end{array}$。
即$\{\begin{array}{l}x - 1 = 2,\\ y - 1 = 3\end{array}$，解得$\{\begin{array}{l}x = 3,\\ y = 4\end{array}$。
$\begin{cases}x = 3\\y = 4\end{cases}$
已知原方程组$\{\begin{array}{l}ax + by = 2,\\ cx + dy = 3\end{array}$的解是$\{\begin{array}{l}x = 2,\\ y = 3\end{array}$，所以新方程组$\{\begin{array}{l}am + bn = 2,\\ cm + dn = 3\end{array}$的解为$\{\begin{array}{l}m = 2,\\ n = 3\end{array}$。
即$\{\begin{array}{l}x - 1 = 2,\\ y - 1 = 3\end{array}$，解得$\{\begin{array}{l}x = 3,\\ y = 4\end{array}$。
$\begin{cases}x = 3\\y = 4\end{cases}$
9. (2024·成都)我国古代数学著作《九章算术》中记载了这样一个题目，其大意如下：今有人合伙买琎石，每人出$\dfrac{1}{2}$钱，会多出$4$钱；每人出$\dfrac{1}{3}$钱，又差了$3$钱.问：人数、琎石价各是多少？设人数为$x$，琎石价为$y$钱，则可列方程组为.
答案：9.$\begin{cases}y = \frac{1}{2}x - 4,\\y = \frac{1}{3}x + 3\end{cases}$
10. 已知关于$x,y$的二元一次方程组$\{\begin{array}{l}4x+3y=1,\\ ax+(a - 1)y=3\end{array} $的解中$x$与$y$的值相等，求$a$的值.
答案：10.由题意，得$x = y$，所以$4x + 3y = 4x + 3x = 1$，解得$x = y = \frac{1}{7}$。将$x = y = \frac{1}{7}$代入$ax + (a - 1)y = 3$，得$\frac{1}{7}a + \frac{1}{7}(a - 1) = 3$，解得$a = 11$
11. 《九章算术》是我国古代数学的经典著作，书中有一个问题：“今有黄金九枚，白银一十一枚，称之重适等，交易其一，金轻十三两，问：金、银各重几何？”大意如下：甲袋中装有黄金$9$枚（每枚黄金的质量相同），乙袋中装有白银$11$枚（每枚白银的质量相同），称重两袋相等，两袋互相交换$1$枚后，甲袋比乙袋轻了$13$两（袋子的质量忽略不计），问：黄金、白银每枚各重多少两（只要求设出适当的未知数，列出方程组，不必求出方程组的解）？
答案：11.设每枚黄金重$x$两，每枚白银重$y$两.根据题意，得$\begin{cases}9x = 11y,\\8x + y = 10y + x - 13\end{cases}$
12. 已知关于$x,y$的二元一次方程组$\{\begin{array}{l}ax+y=b,\\ x - by=a\end{array} $的解是$\{\begin{array}{l}x=1,\\ y=1\end{array} $求$(a + b)^2-(a - b)(a + b)$的值.
答案：12.把$\begin{cases}x = 1,\\y = 1\end{cases}$代入方程组$\begin{cases}ax + y = b,\\x - by = a\end{cases}$得方程组$\begin{cases}a + 1 = b,\\1 - b = a\end{cases}$整理，得$\begin{cases}a - b = - 1,\\a + b = 1\end{cases}$所以$(a + b)^2 - (a - b)(a + b) = 1^2 - (-1) × 1 = 2$