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8. 用代入法解方程组$\begin{cases}4x - 3y = 17 & ①,\\5x + y = 7 & ②,\end{cases}$使得代入后化简比较容易的变形是【 】
A. 由①，得$x = \frac{17 + 3y}{4}$
B. 由①，得$y = \frac{17 - 4x}{-3}$
C. 由②，得$y = 7 - 5x$
D. 由②，得$x = \frac{7 - y}{5}$
A. 由①，得$x = \frac{17 + 3y}{4}$
B. 由①，得$y = \frac{17 - 4x}{-3}$
C. 由②，得$y = 7 - 5x$
D. 由②，得$x = \frac{7 - y}{5}$
答案
8. C
9. 用代入法解方程组$\begin{cases}y = 2x - 1 & ①,\_\_\_\_\_\_ & ②\end{cases}$时，将①代入②，得$x - 4x + 2 = 4$，则方程②可以是【 】
A. $x + 2y = 4$
B. $x - 2y = 4$
C. $2x - y = 4$
D. $2x + y = 4$
A. $x + 2y = 4$
B. $x - 2y = 4$
C. $2x - y = 4$
D. $2x + y = 4$
答案
9. B
10. 已知$\vert 2x + y + 3\vert + (x - y + 3)^2 = 0$，则$(x + y)^{2025} =$【 】
A.$2025$
B.$1$
C.$-1$
D.$-2025$
A.$2025$
B.$1$
C.$-1$
D.$-2025$
答案
10. C
11. 定义一种新运算：$m\odot n = am + bn^2$. 若$3\odot 2 = 10$，$4\odot 1 = 9$，则$1\odot 3 =$
11
.答案
11. 11
12. 解下列方程组：
(1) $\begin{cases}3x + y = 1,\frac{1}{2}x + \frac{1}{4}y = 0;\end{cases}$
(2) $\begin{cases}3x + 4y = 15,\\3x + 6y + 3 = 0;\end{cases}$
(3) $\begin{cases}0.5x + 0.7y = 35,\\x + 0.4y = 40;\end{cases}$
(4) $\begin{cases}2u + 3v = 4,\\4u - 3(u - v) = 11.\end{cases}$
(1) $\begin{cases}3x + y = 1,\frac{1}{2}x + \frac{1}{4}y = 0;\end{cases}$
(2) $\begin{cases}3x + 4y = 15,\\3x + 6y + 3 = 0;\end{cases}$
(3) $\begin{cases}0.5x + 0.7y = 35,\\x + 0.4y = 40;\end{cases}$
(4) $\begin{cases}2u + 3v = 4,\\4u - 3(u - v) = 11.\end{cases}$
答案
12. (1) $ \begin{cases} x = 1, \\ y = -2 \end{cases} $ (2) $ \begin{cases} x = 17, \\ y = -9 \end{cases} $ (3) $ \begin{cases} x = 28, \\ y = 30 \end{cases} $ (4) $ \begin{cases} u = -7, \\ v = 6 \end{cases} $
