4. 解下列方程组：
(1) $\begin{cases}m:n = 2:3, \\ 2 + 5m - 3n = 15\end{cases}$
(2) $\begin{cases}\dfrac{3x + 1}{2026} = \dfrac{4y - 2}{2025}, \\ \dfrac{9x - 2}{3} - \dfrac{8y - 7}{2} = \dfrac{5}{6}\end{cases}$
(1) $\begin{cases}m:n = 2:3, \\ 2 + 5m - 3n = 15\end{cases}$
(2) $\begin{cases}\dfrac{3x + 1}{2026} = \dfrac{4y - 2}{2025}, \\ \dfrac{9x - 2}{3} - \dfrac{8y - 7}{2} = \dfrac{5}{6}\end{cases}$
答案：4. (1) $\begin{cases}m:n = 2:3, &①\\2 + 5m - 3n = 15, &②\end{cases}$ 由①可设 $m = 2k$，$n = 3k$，代入②，得 $2 + 10k - 9k = 15$，所以 $k = 13$。所以 $m = 26$，$n = 39$。所以原方程组的解为 $\begin{cases}m = 26,\ = 39.\end{cases}$
(2) $\begin{cases}\dfrac{3x + 1}{2026} = \dfrac{4y - 2}{2025}, &①\\\dfrac{9x - 2}{3} - \dfrac{8y - 7}{2} = \dfrac{5}{6}, &②\end{cases}$ 设 $\dfrac{3x + 1}{2026} = \dfrac{4y - 2}{2025} = k$，则 $3x = 2026k - 1$，$4y = 2025k + 2$，代入②，得 $(2026k - 1) - \dfrac{2}{3} - (2025k + 2) + \dfrac{7}{2} = \dfrac{5}{6}$，解得 $k = 1$。所以 $x = \dfrac{2026 - 1}{3} = 675$，$y = \dfrac{2025 + 2}{4} = \dfrac{2027}{4}$。所以原方程组的解为 $\begin{cases}x = 675,\\y = \dfrac{2027}{4}.\end{cases}$
(2) $\begin{cases}\dfrac{3x + 1}{2026} = \dfrac{4y - 2}{2025}, &①\\\dfrac{9x - 2}{3} - \dfrac{8y - 7}{2} = \dfrac{5}{6}, &②\end{cases}$ 设 $\dfrac{3x + 1}{2026} = \dfrac{4y - 2}{2025} = k$，则 $3x = 2026k - 1$，$4y = 2025k + 2$，代入②，得 $(2026k - 1) - \dfrac{2}{3} - (2025k + 2) + \dfrac{7}{2} = \dfrac{5}{6}$，解得 $k = 1$。所以 $x = \dfrac{2026 - 1}{3} = 675$，$y = \dfrac{2025 + 2}{4} = \dfrac{2027}{4}$。所以原方程组的解为 $\begin{cases}x = 675,\\y = \dfrac{2027}{4}.\end{cases}$
5. 解下列方程组：
(1) $\begin{cases}\dfrac{x + y}{6} + \dfrac{x - y}{10} = 3, \\ \dfrac{x + y}{6} - \dfrac{x - y}{10} = -1\end{cases}$
(2) $\begin{cases}2^{x} + 3^{y} = 43, \\ 2^{x + 2} - 3^{y} = 37\end{cases}$
(3) $\begin{cases}\dfrac{5m - 2n}{2} - 3 = \dfrac{2m - 5n}{5} + 1, \\ \dfrac{3(5m - 2n)}{2} = \dfrac{7(2m - 5n)}{5} + 28\end{cases}$
(1) $\begin{cases}\dfrac{x + y}{6} + \dfrac{x - y}{10} = 3, \\ \dfrac{x + y}{6} - \dfrac{x - y}{10} = -1\end{cases}$
(2) $\begin{cases}2^{x} + 3^{y} = 43, \\ 2^{x + 2} - 3^{y} = 37\end{cases}$
(3) $\begin{cases}\dfrac{5m - 2n}{2} - 3 = \dfrac{2m - 5n}{5} + 1, \\ \dfrac{3(5m - 2n)}{2} = \dfrac{7(2m - 5n)}{5} + 28\end{cases}$
答案：5. (1) 把 $\dfrac{x + y}{6}$，$\dfrac{x - y}{10}$ 分别看作一个整体，分别设为 $m$，$n$，原方程组化为 $\begin{cases}m + n = 3,\\m - n = -1,\end{cases}$ 解得 $\begin{cases}m = 1,\ = 2,\end{cases}$ 所以 $\begin{cases}\dfrac{x + y}{6} = 1,\\\dfrac{x - y}{10} = 2,\end{cases}$ 解得 $\begin{cases}x = 13,\\y = -7.\end{cases}$ 所以原方程组的解为 $\begin{cases}x = 13,\\y = -7.\end{cases}$
(2) 设 $2^x = m$，$3^y = n$，原方程组可化为 $\begin{cases}m + n = 43,\\4m - n = 37,\end{cases}$ 解得 $\begin{cases}m = 16,\ = 27,\end{cases}$ 即 $\begin{cases}2^x = 16,\\3^y = 27,\end{cases}$ 解得 $\begin{cases}x = 4,\\y = 3,\end{cases}$ 故原方程组的解为 $\begin{cases}x = 4,\\y = 3.\end{cases}$
(3) 将 $\dfrac{5m - 2n}{2}$，$\dfrac{2m - 5n}{5}$ 分别看作一个整体，分别设为 $u$，$v$，得 $\begin{cases}u - 3 = v + 1,\\3u = 7v + 28,\end{cases}$ 解得 $\begin{cases}u = 0,\\v = -4,\end{cases}$ 即 $\begin{cases}\dfrac{5m - 2n}{2} = 0,\\\dfrac{2m - 5n}{5} = -4,\end{cases}$ 解得 $\begin{cases}m = \dfrac{40}{21},\ = \dfrac{100}{21},\end{cases}$ 所以原方程组的解为 $\begin{cases}m = \dfrac{40}{21},\ = \dfrac{100}{21}.\end{cases}$
(2) 设 $2^x = m$，$3^y = n$，原方程组可化为 $\begin{cases}m + n = 43,\\4m - n = 37,\end{cases}$ 解得 $\begin{cases}m = 16,\ = 27,\end{cases}$ 即 $\begin{cases}2^x = 16,\\3^y = 27,\end{cases}$ 解得 $\begin{cases}x = 4,\\y = 3,\end{cases}$ 故原方程组的解为 $\begin{cases}x = 4,\\y = 3.\end{cases}$
(3) 将 $\dfrac{5m - 2n}{2}$，$\dfrac{2m - 5n}{5}$ 分别看作一个整体，分别设为 $u$，$v$，得 $\begin{cases}u - 3 = v + 1,\\3u = 7v + 28,\end{cases}$ 解得 $\begin{cases}u = 0,\\v = -4,\end{cases}$ 即 $\begin{cases}\dfrac{5m - 2n}{2} = 0,\\\dfrac{2m - 5n}{5} = -4,\end{cases}$ 解得 $\begin{cases}m = \dfrac{40}{21},\ = \dfrac{100}{21},\end{cases}$ 所以原方程组的解为 $\begin{cases}m = \dfrac{40}{21},\ = \dfrac{100}{21}.\end{cases}$