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5. 已知方程组 $\begin{cases}ax + y = 3,\\3x - 2y = 5\end{cases}$ 的解 $x$ 和 $y$ 互为相反数，求 $a$ 的值.
答案
由题意，得$x + y = 0$，即$x = - y$，把$x = - y$代入方程组，得$\{\begin{array}{l}-ay + y = 3\\-3y - 2y = 5\end{array} $，解得$\{\begin{array}{l}a = 4\\y = - 1\end{array} $.
∴$a$的值为$4$.
∴$a$的值为$4$.
6. 已知关于 $x$、$y$ 的二元一次方程组 $\begin{cases}x - 2y = k,\\2x + 3y = k + 1\end{cases}$ 的解的和为 $1$，求 $k$ 的值.
答案
解原方程组，得$\{\begin{array}{l}x = \frac{5k + 2}{7}\\y = \frac{1 - k}{7}\end{array} $.
∵原方程组的解的和为$1$，
∴$\frac{5k + 2}{7} + \frac{1 - k}{7} = 1$，解得$k = 1$.
∵原方程组的解的和为$1$，
∴$\frac{5k + 2}{7} + \frac{1 - k}{7} = 1$，解得$k = 1$.
7. 已知关于 $x$、$y$ 的二元一次方程组 $\begin{cases}2x - 3y = 7a - 9,\\x + 2y = -1\end{cases}$ 的解满足方程 $2x - y = 13$，求 $a$ 的值.
答案
由题意，得$\{\begin{array}{l}x + 2y = - 1\\2x - y = 13\end{array} $，解得$\{\begin{array}{l}x = 5\\y = - 3\end{array} $. 将$\{\begin{array}{l}x = 5\\y = - 3\end{array} $代入$2x - 3y = 7a - 9$，得$2×5 - 3×(-3) = 7a - 9$，解得$a = 4$.
8. 已知方程组 $\begin{cases}2x - y = -3,\\ax + 5y = 4\end{cases}$ 与 $\begin{cases}x - y = 3,\\5x + by = 1\end{cases}$ 有相同的解，求 $a$、$b$ 的值.
答案
由题意，得$\{\begin{array}{l}2x - y = - 3\\x - y = 3\end{array} $，解得$\{\begin{array}{l}x = - 6\\y = - 9\end{array} $. 将$\{\begin{array}{l}x = - 6\\y = - 9\end{array} $代入$\{\begin{array}{l}ax + 5y = 4\\5x + by = 1\end{array} $，得$\{\begin{array}{l}-6a - 45 = 4\\-30 - 9b = 1\end{array} $，解得$\{\begin{array}{l}a = -\frac{49}{6}\\b = -\frac{31}{9}\end{array} $.
解析
解：由题意，得
$\begin{cases}2x - y = -3 \\x - y = 3\end{cases}$
解得
$\begin{cases}x = -6 \\y = -9\end{cases}$
将$\begin{cases}x = -6 \\ y = -9\end{cases}$代入$\begin{cases}ax + 5y = 4 \\ 5x + by = 1\end{cases}$，得
$\begin{cases}-6a + 5×(-9) = 4 \\5×(-6) + b×(-9) = 1\end{cases}$
即
$\begin{cases}-6a - 45 = 4 \\-30 - 9b = 1\end{cases}$
解得
$\begin{cases}a = -\dfrac{49}{6} \\b = -\dfrac{31}{9}\end{cases}$
$\begin{cases}2x - y = -3 \\x - y = 3\end{cases}$
解得
$\begin{cases}x = -6 \\y = -9\end{cases}$
将$\begin{cases}x = -6 \\ y = -9\end{cases}$代入$\begin{cases}ax + 5y = 4 \\ 5x + by = 1\end{cases}$，得
$\begin{cases}-6a + 5×(-9) = 4 \\5×(-6) + b×(-9) = 1\end{cases}$
即
$\begin{cases}-6a - 45 = 4 \\-30 - 9b = 1\end{cases}$
解得
$\begin{cases}a = -\dfrac{49}{6} \\b = -\dfrac{31}{9}\end{cases}$
9. 已知方程组 $\begin{cases}2x + 5y = -6,\\ax - by = -4\end{cases}$ 与方程组 $\begin{cases}3x - 5y = 16,\\bx + ay = -8\end{cases}$ 的解相同，求 $(2a + b)^{2025}$ 的值.
答案
由题意，得$\{\begin{array}{l}2x + 5y = - 6\\3x - 5y = 16\end{array} $，解得$\{\begin{array}{l}x = 2\\y = - 2\end{array} $. 将$\{\begin{array}{l}x = 2\\y = - 2\end{array} $代入$\{\begin{array}{l}ax - by = - 4\\bx + ay = - 8\end{array} $，得$\{\begin{array}{l}2a + 2b = - 4\\2b - 2a = - 8\end{array} $，解得$\{\begin{array}{l}a = 1\\b = - 3\end{array} $，
∴$(2a + b)^{2026} = [2×1 + (-3)]^{2026} = (-1)^{2026} = 1$.
∴$(2a + b)^{2026} = [2×1 + (-3)]^{2026} = (-1)^{2026} = 1$.
