1. 用代入法解下列方程组：
(1) $\begin{cases} x = 3y, \\ x + 4y = 14; \end{cases}$
(2) $\begin{cases} y = 2x - 3, \\ 3x + 2y = 8; \end{cases}$
(3) $\begin{cases} x - y = 1, \\ 4(x - y) - y = 5; \end{cases}$
(4) $\begin{cases} x - 2 = 2(y - 1), \\ 2(x - 2) + y - 1 = 5. \end{cases}$

(1)$\{\begin{array}{l} x=3y①,\\ x+4y=14②.\end{array} $把①代入②，得$3y + 4y = 14$，解得$y = 2$。把$y = 2$代入①，得$x = 6$。
∴原方程组的解为$\{\begin{array}{l} x=6,\\ y=2.\end{array} $
(2)$\{\begin{array}{l} y=2x - 3①,\\ 3x + 2y = 8②.\end{array} $把①代入②，得$3x + 2(2x - 3) = 8$，解得$x = 2$。把$x = 2$代入①，得$y = 1$。
∴原方程组的解为$\{\begin{array}{l} x=2,\\ y=1.\end{array} $
(3)$\{\begin{array}{l} x - y = 1①,\\ 4(x - y) - y = 5②.\end{array} $把①代入②，得$4 - y = 5$，解得$y = - 1$。把$y = - 1$代入①，得$x - (-1) = 1$，解得$x = 0$。
∴原方程组的解为$\{\begin{array}{l} x=0,\\ y=-1.\end{array} $
(4)$\{\begin{array}{l} x - 2 = 2(y - 1)①,\\ 2(x - 2) + y - 1 = 5②.\end{array} $把①代入②，得$2×2(y - 1) + y - 1 = 5$，解得$y = 2$。把$y = 2$代入①，得$x - 2 = 2×(2 - 1)$，$x = 4$。
∴原方程组的解为$\{\begin{array}{l} x=4,\\ y=2.\end{array} $

2. 用代入法解下列方程组：

(1) $\begin{cases} 3x = 5y, \\ 2x - 3y = 1; \end{cases}$
(2) $\begin{cases} 3s - t = 5, \\ 5s + 2t = 15; \end{cases}$

(3) $\begin{cases} 2x = 5(x + y), \\ 3x - 10(x + y) = 2; \end{cases}$
(4) $\begin{cases} \dfrac{m}{6} + \dfrac{n}{3} = 13, \\ \dfrac{m}{3} - \dfrac{n}{4} = \dfrac{37}{12}. \end{cases}$

(1)$\{\begin{array}{l} 3x = 5y①,\\ 2x - 3y = 1②.\end{array} $由①，得$x = \frac{5y}{3}$③，将③代入②，得$2×\frac{5y}{3} - 3y = 1$，解得$y = 3$。将$y = 3$代入③，得$x = 5$。
∴原方程组的解为$\{\begin{array}{l} x=5,\\ y=3.\end{array} $
(2)$\{\begin{array}{l} 3s - t = 5①,\\ 5s + 2t = 15②.\end{array} $由①，得$t = 3s - 5$③，把③代入②，得$5s + 2(3s - 5) = 15$，解得$s = \frac{25}{11}$。把$s = \frac{25}{11}$代入③，得$t = \frac{20}{11}$。
∴原方程组的解为$\{\begin{array}{l} s=\frac{25}{11},\\ t=\frac{20}{11}.\end{array} $
(3)$\{\begin{array}{l} 2x = 5(x + y)①,\\ 3x - 10(x + y) = 2②.\end{array} $把①代入②，得$3x - 2×2x = 2$，解得$x = - 2$。把$x = - 2$代入①，得$2×(-2) = 5×(-2 + y)$，解得$y = \frac{6}{5}$。
∴原方程组的解为$\{\begin{array}{l} x=-2,\\ y=\frac{6}{5}.\end{array} $
(4)$\{\begin{array}{l} \frac{m}{6} + \frac{n}{3} = 13①,\\ \frac{m}{3} - \frac{n}{4} = \frac{37}{12}②.\end{array} $由②，得$\frac{m}{3} = \frac{37}{12} + \frac{n}{4}$③，将③代入①，得$\frac{1}{2}(\frac{37}{12} + \frac{n}{4}) + \frac{n}{3} = 13$，解得$n = 25$。将$n = 25$代入③，得$\frac{m}{3} = \frac{37}{12} + \frac{25}{4}$，解得$m = 28$。
∴原方程组的解为$\{\begin{array}{l} m=28,\\ n=25.\end{array} $