13. 解下列方程组：
(1)$\begin{cases}\dfrac{1 - 2s}{5} = \dfrac{t}{6},\\1 - \dfrac{t - s}{4} = \dfrac{s}{2};\end{cases}$
(2)$\begin{cases}\dfrac{m + n}{3} - \dfrac{n - m}{4} = 2,\\4m + \dfrac{n}{3} = 14;\end{cases}$
(3)$\begin{cases}\dfrac{x}{3} + \dfrac{y}{5} = 1,\\3(x + y) + 2(x - 3y) = 15;\end{cases}$
(4)$\begin{cases}2(x + 3) = 3 - 5(y - 2),\\\dfrac{x + 1}{3} - \dfrac{2y + 1}{2} = 1.\end{cases}$

13. (1) $\begin{cases}s=-2,\\t=6\end{cases}$ (2) $\begin{cases}m=3\dfrac{3}{5},\\n=-1\dfrac{1}{5}\end{cases}$ (3) $\begin{cases}x=3,\\y=0\end{cases}$ (4) $\begin{cases}x=3.5,\\y=0\end{cases}$

14. (济宁期末)阅读材料：
善于思考的乐乐同学在解方程组$\begin{cases}3(m + 5) - 2(n + 3) = -1,\\3(m + 5) + 2(n + 3) = 7\end{cases}$时，采用了一种“整体换元”的解法。即：把$m + 5$、$n + 3$看成一个整体，设$m + 5 = x$，$n + 3 = y$，则原方程组可化为$\begin{cases}3x - 2y = -1,\\3x + 2y = 7,\end{cases}$解得$\begin{cases}x = 1,\\y = 2,\end{cases}$即$\begin{cases}m + 5 = 1,\\n + 3 = 2,\end{cases}$解得$\begin{cases}m = -4,\\n = -1.\end{cases}$
(1) 学以致用：模仿乐乐同学的“整体换元”的方法，解方程组$\begin{cases}\dfrac{x + y}{3} + \dfrac{x - y}{5} = 4,\\\dfrac{x + y}{3} - \dfrac{x - y}{5} = -2.\end{cases}$
(2) 拓展提升：已知关于$x$、$y$的方程组$\begin{cases}a_1x - b_1y = c_1,\\a_2x - b_2y = c_2\end{cases}$的解为$\begin{cases}x = 3,\\y = 4,\end{cases}$请直接写出关于$m$、$n$的方程组$\begin{cases}a_1(m + 2) - b_1n = c_1,\\a_2(m + 2) - b_2n = c_2\end{cases}$的解： ______ 。

14. (1) 令 $m=\dfrac{x+y}{3}$，$n=\dfrac{x-y}{5}$，则原方程组可化为 $\begin{cases}m+n=4,\\m-n=-2,\end{cases}$ 解得 $\begin{cases}m=1,\\n=3.\end{cases}$ $\therefore$ $\begin{cases}\dfrac{x+y}{3}=1,\\\dfrac{x-y}{5}=3,\end{cases}$ 解得 $\begin{cases}x=9,\\y=-6\end{cases}$ (2) $\begin{cases}m=1,\\n=4\end{cases}$