4. 若$\begin{cases}x = 2,\\y = -1\end{cases}$是关于$x$、$y$的二元一次方程$ax + by - 5 = 0$的一组解，则$4a - 2b - 9$的值为 ______ 。

4. 1 解析:把$\{\begin{array}{l} x=2,\\ y=-1\end{array} $代入方程$ax+by-5=0$,得$2a-b-5=0$,移项,得$2a-b=5,$$\therefore 4a-2b-9=2(2a-b)-9=2×5-9=10-9=1.$

5. 如果关于$x$、$y$的二元一次方程组$\begin{cases}a_1x + b_1y = c_1,\\a_2x + b_2y = c_2\end{cases}$的解是$\begin{cases}x = 2,\\y = 3,\end{cases}$那么关于$x$、$y$的二元一次方程组$\begin{cases}2a_1x + b_1y = c_1 + 3a_1,\\2a_2x + b_2y = c_2 + 3a_2\end{cases}$的解是 ______ 。

5. $\{\begin{array}{l} x=\frac {5}{2},\\ y=3\end{array} $ 解析:将方程组$\{\begin{array}{l} 2a_{1}x+b_{1}y=c_{1}+3a_{1},\\ 2a_{2}x+b_{2}y=c_{2}+3a_{2}\end{array} $变形为$\{\begin{array}{l} (2x-3)a_{1}+b_{1}y=c_{1},\\ (2x-3)a_{2}+b_{2}y=c_{2},\end{array} $根据题意,得$\{\begin{array}{l} 2x-3=2,\\ y=3,\end{array} $解得$\{\begin{array}{l} x=\frac {5}{2},\\ y=3.\end{array} $

将方程组$\begin{cases}2a_1x + b_1y = c_1 + 3a_1\\2a_2x + b_2y = c_2 + 3a_2\end{cases}$变形为$\begin{cases}(2x - 3)a_1 + b_1y = c_1\\(2x - 3)a_2 + b_2y = c_2\end{cases}$，因为关于$x$、$y$的二元一次方程组$\begin{cases}a_1x + b_1y = c_1\\a_2x + b_2y = c_2\end{cases}$的解是$\begin{cases}x = 2\\y = 3\end{cases}$，所以可得$\begin{cases}2x - 3 = 2\\y = 3\end{cases}$，解得$\begin{cases}x = \frac{5}{2}\\y = 3\end{cases}$。
$\begin{cases} x=\dfrac{5}{2} \\ y=3 \end{cases}$

6. 已知方程组$\begin{cases}ax + by = 3,\\5x - cy = 1,\end{cases}$甲正确地解得$\begin{cases}x = 2,\\y = 3,\end{cases}$而乙粗心地把$c$看错了，解得$\begin{cases}x = 3,\\y = 6,\end{cases}$试求出$a$、$b$、$c$的值。

6. 根据题意,得$\{\begin{array}{l} 2a+3b=3,\\ 3a+6b=3,\end{array} $解得$\{\begin{array}{l} a=3,\\ b=-1,\end{array} $把$\{\begin{array}{l} x=2,\\ y=3\end{array} $代入方程$5x-cy=1$,得$10-3c=1$,解得$c=3.$

解：将$\begin{cases}x = 2\\y = 3\end{cases}$代入$ax + by = 3$，得$2a + 3b = 3$；
将$\begin{cases}x = 3\\y = 6\end{cases}$代入$ax + by = 3$，得$3a + 6b = 3$；
联立方程组$\begin{cases}2a + 3b = 3\\3a + 6b = 3\end{cases}$，
由$3a + 6b = 3$化简得$a + 2b = 1$，即$a = 1 - 2b$，
将$a = 1 - 2b$代入$2a + 3b = 3$，得$2(1 - 2b) + 3b = 3$，
解得$b = -1$，则$a = 1 - 2×(-1) = 3$；
将$\begin{cases}x = 2\\y = 3\end{cases}$代入$5x - cy = 1$，得$10 - 3c = 1$，解得$c = 3$；
综上，$a = 3$，$b = -1$，$c = 3$。

7. 用加减消元法解二元一次方程组$\begin{cases}3x - y = 5①,\\5x + 2y = 15②\end{cases}$时，下列做法中无法消元的是( )

A.$①×2 + ②$
B.$①×5 - ②×3$
C.$①×3 - ②×5$
D.$①×(-5) + ②×3$

7. C 解析:①$×2+$②,得$11x=25$,能消元,故A选项不符合题意;①$×5-$②$×3$,得$-11y=-20$,能消元,故B选项不符合题意;①$×3-$②$×5$,得$-16x-13y=-60$,不能消元,故C选项符合题意;①$×(-5)+$②$×3$,得$11y=20$,能消元,故D选项不符合题意.

8. 若$(a + b - 1)^2 + |2a - b + 7| = 0$，则$a^b =$。

8. -8 解析:$\because (a+b-1)^{2}+|2a-b+7|=0,\therefore \{\begin{array}{l} a+b-1=0,\\ 2a-b+7=0,\end{array} $解得$\{\begin{array}{l} a=-2,\\ b=3,\end{array} $$\therefore a^{b}=(-2)^{3}=-8$.

9. 解下列方程组：
(1) $\begin{cases}x - 2y = 7,\\x + y = 10;\end{cases}$
(2) $\begin{cases}3x + 2y = 10,\\\dfrac{x}{2} = 1 + \dfrac{y + 1}{3}.\end{cases}$

9. (1)$\{\begin{array}{l} x-2y=7①,\\ x+y=10②.\end{array} $由①,得$x=2y+7$③,把③代入②,得$2y+7+y=10$,解得$y=1$.把$y=1$代入③,得$x=9.\therefore $原方程组的解是$\{\begin{array}{l} x=9,\\ y=1.\end{array} $ (2)原方程组整理,得$\{\begin{array}{l} 3x+2y=10①,\\ 3x-2y=8②.\end{array} $①+②,得$6x=18$,解得$x=3$.把$x=3$代入①,得$3×3+2y=10$,解得$y=\frac {1}{2}.\therefore $原方程组的解是$\{\begin{array}{l} x=3,\\ y=\frac {1}{2}.\end{array} $

10. 解关于$x$、$y$方程组$\begin{cases}(m + 1)x - (3n + 2)y = 8①,\\(5 - n)x + my = 11②,\end{cases}$可以用$①×2 + ②$消去未知数$x$，也可以用$① + ②×5$消去未知数$y$，求$m$、$n$的值。

10. 根据题意,得$\{\begin{array}{l} 2(m+1)+(5-n)=0,\\ -(3n+2)+5m=0,\end{array} $整理,得$\{\begin{array}{l} 2m-n=-7,\\ 5m-3n=2,\end{array} $解得$\{\begin{array}{l} m=-23,\\ n=-39.\end{array} $

解：根据题意，得
$\begin{cases}2(m + 1) + (5 - n) = 0 \\-(3n + 2) + 5m = 0\end{cases}$
整理，得
$\begin{cases}2m - n = -7 \\5m - 3n = 2\end{cases}$
解得
$\begin{cases}m = -23 \ = -39\end{cases}$