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解：$\begin{cases} 3x + 4y = 18 & \textcircled{1} \\ \frac{1}{2}x + y = 4 & \textcircled{2} \end{cases}$
由$\textcircled{2}$得：$y = 4 - \frac{1}{2}x$ ③
将③代入①，得$3x + 4(4 - \frac{1}{2}x) = 18$
$3x + 16 - 2x = 18$
解得$x = 2$
把$x = 2$代入③，得$y = 4 - \frac{1}{2} × 2 = 3$
所以方程组的解为$\begin{cases} x=2 \\ y=3 \end{cases}$
解：$\begin{cases} x - 2y = 5 & \textcircled{1} \\ 3x + 4y = 25 & \textcircled{2} \end{cases}$
$\textcircled{1} × 2$得：$2x - 4y = 10$ ③
$\textcircled{2} + \textcircled{3}$得：$5x = 35，$解得$x = 7$
把$x = 7$代入①，得$7 - 2y = 5，$解得$y = 1$
所以方程组的解为$\begin{cases} x=7 \\ y=1 \end{cases}$
解：$\begin{cases} 3x - 2y = 7 & \textcircled{1} \\ \frac{x - 2}{3} - \frac{2y - 1}{2} = 1 & \textcircled{2} \end{cases}$
整理②：$2(x - 2) - 3(2y - 1) = 6$
$2x - 4 - 6y + 3 = 6$
$2x - 6y = 7$ ③
$\textcircled{1} × 3$得：$9x - 6y = 21$ ④
$\textcircled{4} - \textcircled{3}$得：$7x = 14，$解得$x = 2$
把$x = 2$代入①，得$6 - 2y = 7，$解得$y = -\frac{1}{2}$
所以方程组的解为$\begin{cases} x=2 \\ y=-\frac{1}{2} \end{cases}$
解：$\begin{cases} x + y = 3 & \textcircled{1} \\ z + x = 7 & \textcircled{2} \\ y + z = -2 & \textcircled{3} \end{cases}$
$\textcircled{1} + \textcircled{2} + \textcircled{3}$得：$2x + 2y + 2z = 8，$即$x + y + z = 4$ ④
$\textcircled{4} - \textcircled{1}$得：$z = 1$
$\textcircled{4} - \textcircled{2}$得：$y = -3$
$\textcircled{4} - \textcircled{3}$得：$x = 6$
所以方程组的解为$\begin{cases} x=6 \\ y=-3 \\ z=1 \end{cases}$
解：$\begin{cases} x + 2y = k - 1 & \textcircled{1} \\ 2x + y = 5k + 4 & \textcircled{2} \end{cases}$
$\textcircled{1} + \textcircled{2}$得：$3x + 3y = 6k + 3，$即$x + y = 2k + 1$
因为$x + y = 5，$所以$2k + 1 = 5$
解得$k = 2$
解：上述解答过程不正确，从第①步开始出现错误。
正确解答过程：
$(x^2 + ax + b)(2x^2 - 3x - 1)$
$= 2x^4 - 3x^3 - x^2 + 2ax^3 - 3ax^2 - ax + 2bx^2 - 3bx - b$
$= 2x^4 + (-3 + 2a)x^3 + (-1 - 3a + 2b)x^2 + (-a - 3b)x - b$
根据题意，得$\begin{cases} -3 + 2a = -5 \\ -1 - 3a + 2b = -6 \end{cases}$
解得$\begin{cases} a = -1 \\ b = -4 \end{cases}$
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