7. (常州期末)现有方程组$\begin{cases}x - y = m,\\2x + 3y = 3m + 1,\end{cases}$消去$m$，得$x$与$y$的关系式为( )

A.$3x + 2y = 1$
B.$x + 4y = 1$
C.$5x + 6y = 1$
D.$x - 6y = - 1$

7. D

由$x - y = m$得$m = x - y$，代入$2x + 3y = 3m + 1$，
得$2x + 3y = 3(x - y) + 1$，
$2x + 3y = 3x - 3y + 1$，
$2x + 3y - 3x + 3y = 1$，
$-x + 6y = 1$，
$x - 6y = -1$。
D

8. 若$(2x - y)^2$与$\vert x + 2y - 5\vert$互为相反数，则$(x - y)^{2026}=$。

8. 1

因为$(2x - y)^2$与$\vert x + 2y - 5\vert$互为相反数，所以$(2x - y)^2 + \vert x + 2y - 5\vert = 0$。
由于$(2x - y)^2 ≥ 0$，$\vert x + 2y - 5\vert ≥ 0$，则有：
$\begin{cases}2x - y = 0 \\x + 2y - 5 = 0\end{cases}$
由$2x - y = 0$得$y = 2x$，代入$x + 2y - 5 = 0$：
$x + 2(2x) - 5 = 0 \implies x + 4x = 5 \implies 5x = 5 \implies x = 1$
则$y = 2x = 2×1 = 2$。
所以$x - y = 1 - 2 = -1$，$(x - y)^{2026} = (-1)^{2026} = 1$。
1

9. 若关于$x$，$y$的二元一次方程组$\begin{cases}2x + y = 3m,\\x + 2y = - 3\end{cases}$的解满足$x + y = 1$，则$m$的值是 ______ 。

9. 2

解：$\begin{cases}2x + y = 3m,①\\x + 2y = - 3,②\end{cases}$
①+②，得$3x + 3y = 3m - 3$，
两边同时除以3，得$x + y = m - 1$。
因为$x + y = 1$，所以$m - 1 = 1$，解得$m = 2$。
2

10. 在代数式$ax + by$中，当$\begin{cases}x = 1,\\y = 1\end{cases}$时，它的值是$-6$；当$\begin{cases}x = 2,\\y = 3\end{cases}$时，它的值是$3$。当$\begin{cases}x = 1,\\y = - 1\end{cases}$时，代数式$ax + by$的值为 ______ 。

10. -36

由题意得：
$\begin{cases}a + b = -6 \\2a + 3b = 3\end{cases}$
由第一个方程得：$a = -6 - b$，代入第二个方程：
$2(-6 - b) + 3b = 3$
$-12 - 2b + 3b = 3$
$b = 15$
则$a = -6 - 15 = -21$
当$x = 1$，$y = -1$时，$ax + by = -21×1 + 15×(-1) = -21 - 15 = -36$
-36

11. 用代入法解下列方程组：
(1)$\begin{cases}2x + y = 11,\\3x + y = 12;\end{cases}$
(2)$\begin{cases}2x + y = 1,\\4x - y = 5;\end{cases}$
(3)$\begin{cases}x + 3y = - 1,\\3x - 2y = 8;\end{cases}$
(4)$\begin{cases}x - y = 5,\\2x + y = 4;\end{cases}$
(5)$\begin{cases}3x + 2y = 5x + 2,\\2(3x + 2y)=2x + 8;\end{cases}$
(6)$\begin{cases}2a + 3b = 4,\\4a - 9b = - 1.\end{cases}$

11. (1)解：由$2x + y = 11$得$y = 11 - 2x$，代入$3x + y = 12$，得$3x + 11 - 2x = 12$，解得$x = 1$，将$x = 1$代入$y = 11 - 2x$，得$y = 9$，所以$\begin{cases}x = 1\\y = 9\end{cases}$
(2)解：由$2x + y = 1$得$y = 1 - 2x$，代入$4x - y = 5$，得$4x - (1 - 2x) = 5$，解得$x = 1$，将$x = 1$代入$y = 1 - 2x$，得$y = -1$，所以$\begin{cases}x = 1\\y = -1\end{cases}$
(3)解：由$x + 3y = -1$得$x = -1 - 3y$，代入$3x - 2y = 8$，得$3(-1 - 3y) - 2y = 8$，解得$y = -1$，将$y = -1$代入$x = -1 - 3y$，得$x = 2$，所以$\begin{cases}x = 2\\y = -1\end{cases}$
(4)解：由$x - y = 5$得$x = y + 5$，代入$2x + y = 4$，得$2(y + 5) + y = 4$，解得$y = -2$，将$y = -2$代入$x = y + 5$，得$x = 3$，所以$\begin{cases}x = 3\\y = -2\end{cases}$
(5)解：由$3x + 2y = 5x + 2$得$3x + 2y = 5x + 2$，代入$2(3x + 2y) = 2x + 8$，得$2(5x + 2) = 2x + 8$，解得$x = \dfrac{1}{2}$，将$x = \dfrac{1}{2}$代入$3x + 2y = 5x + 2$，得$y = \dfrac{3}{2}$，所以$\begin{cases}x = \dfrac{1}{2}\\y = \dfrac{3}{2}\end{cases}$
(6)解：由$2a + 3b = 4$得$2a = 4 - 3b$，即$a = \dfrac{4 - 3b}{2}$，代入$4a - 9b = -1$，得$4×\dfrac{4 - 3b}{2} - 9b = -1$，解得$b = \dfrac{3}{5}$，将$b = \dfrac{3}{5}$代入$a = \dfrac{4 - 3b}{2}$，得$a = \dfrac{11}{10}$，所以$\begin{cases}a = \dfrac{11}{10}\\b = \dfrac{3}{5}\end{cases}$

(1)解：由$2x + y = 11$得$y = 11 - 2x$，代入$3x + y = 12$，得$3x + 11 - 2x = 12$，解得$x = 1$，将$x = 1$代入$y = 11 - 2x$，得$y = 9$，所以$\begin{cases}x = 1\\y = 9\end{cases}$
(2)解：由$2x + y = 1$得$y = 1 - 2x$，代入$4x - y = 5$，得$4x - (1 - 2x) = 5$，解得$x = 1$，将$x = 1$代入$y = 1 - 2x$，得$y = -1$，所以$\begin{cases}x = 1\\y = -1\end{cases}$
(3)解：由$x + 3y = -1$得$x = -1 - 3y$，代入$3x - 2y = 8$，得$3(-1 - 3y) - 2y = 8$，解得$y = -1$，将$y = -1$代入$x = -1 - 3y$，得$x = 2$，所以$\begin{cases}x = 2\\y = -1\end{cases}$
(4)解：由$x - y = 5$得$x = y + 5$，代入$2x + y = 4$，得$2(y + 5) + y = 4$，解得$y = -2$，将$y = -2$代入$x = y + 5$，得$x = 3$，所以$\begin{cases}x = 3\\y = -2\end{cases}$
(5)解：由$3x + 2y = 5x + 2$得$3x + 2y = 5x + 2$，代入$2(3x + 2y) = 2x + 8$，得$2(5x + 2) = 2x + 8$，解得$x = \dfrac{1}{2}$，将$x = \dfrac{1}{2}$代入$3x + 2y = 5x + 2$，得$y = \dfrac{3}{2}$，所以$\begin{cases}x = \dfrac{1}{2}\\y = \dfrac{3}{2}\end{cases}$
(6)解：由$2a + 3b = 4$得$2a = 4 - 3b$，即$a = \dfrac{4 - 3b}{2}$，代入$4a - 9b = -1$，得$4×\dfrac{4 - 3b}{2} - 9b = -1$，解得$b = \dfrac{3}{5}$，将$b = \dfrac{3}{5}$代入$a = \dfrac{4 - 3b}{2}$，得$a = \dfrac{11}{10}$，所以$\begin{cases}a = \dfrac{11}{10}\\b = \dfrac{3}{5}\end{cases}$

12. 对整数$x$，$y$定义一种新运算$T$，规定$T(x,y)=ax^y - by^x$(其中$a$，$b$是常数)，如$T(2,1)=a×2^1 - b×1^2 = 2a - b$。
(1) 填空：$T(2,-1)=$。(用含$a$，$b$的代数式表示)
(2) 若$T(3,2)=10$，$T(8,-1)=-\frac{3}{4}$。
① 求$a$与$b$的值；
② 若$T(x,1)=T(1,x)$，求此时$x$的值。

12. (1)$\dfrac{1}{2}a - b$
(2)解:①由题意,得$\begin{cases}9a - 8b = 10,\\\dfrac{1}{8}a - b = -\dfrac{3}{4},\end{cases}$解得$\begin{cases}a = 2,\\b = 1.\end{cases}$
即$a$的值是 2,$b$的值是 1.
②由题意,得$2x - 1 = 2 - x$,解得$x = 1$.