答案：1. (1) $\begin{cases}x - y = 3, &①\\2y + 3(x - y) = 11, &②\end{cases}$ 将①代入②，得 $2y + 3×3 = 11$，解得 $y = 1$。把 $y = 1$ 代入①，得 $x = 4$。所以原方程组的解为 $\begin{cases}x = 4,\\y = 1.\end{cases}$
(2) $\begin{cases}2x - 3y - 2 = 0, &①\\\dfrac{2x - 3y + 5}{7} + 2y = 9, &②\end{cases}$ 由①得 $2x - 3y = 2$ ③，把③代入②，得 $\dfrac{2 + 5}{7} + 2y = 9$，解得 $y = 4$。把 $y = 4$ 代入①，得 $2x - 3×4 - 2 = 0$，解得 $x = 7$。所以原方程组的解为 $\begin{cases}x = 7,\\y = 4.\end{cases}$
(3) $\begin{cases}4x - 2y = 3, &①\\(3x - y)(2x - y) = 4, &②\end{cases}$ 由①得 $2x - y = \dfrac{3}{2}$ ③，把③代入②，得 $(x + \dfrac{3}{2})×\dfrac{3}{2} = 4$，解得 $x = \dfrac{7}{6}$。把 $x = \dfrac{7}{6}$ 代入③，得 $2×\dfrac{7}{6} - y = \dfrac{3}{2}$，解得 $y = \dfrac{5}{6}$，所以原方程组的解为 $\begin{cases}x = \dfrac{7}{6},\\y = \dfrac{5}{6}.\end{cases}$

答案：2. (1) $\begin{cases}57x + 65y = 179, &①\\65x + 57y = 187, &②\end{cases}$ 由①+②，得 $122x + 122y = 366$，即 $x + y = 3$ ③，②-①，得 $8x - 8y = 8$，即 $x - y = 1$ ④。③+④，得 $2x = 4$，解得 $x = 2$。③-④，得 $2y = 2$，解得 $y = 1$。所以原方程组的解为 $\begin{cases}x = 2,\\y = 1.\end{cases}$
(2) $\begin{cases}8359x + 1641y = 28359, &①\\1641x + 8359y = 21641, &②\end{cases}$ 由①-②，得 $6718x - 6718y = 6718$，即 $x - y = 1$。由①+②，得 $10000x + 10000y = 50000$，即 $x + y = 5$。将它们组成新方程组，得 $\begin{cases}x - y = 1,\\x + y = 5,\end{cases}$ 解得 $\begin{cases}x = 3,\\y = 2.\end{cases}$ 所以原方程组的解为 $\begin{cases}x = 3,\\y = 2.\end{cases}$

答案：3. 原方程组可化为 $\begin{cases}\dfrac{x}{3} - \dfrac{y}{15} = \dfrac{4}{3}, &①\\\dfrac{x}{2} - \dfrac{y}{5} = \dfrac{4}{3}, &②\end{cases}$ ②-①得 $\dfrac{x}{6} - \dfrac{2y}{15} = 0$，即 $\dfrac{x}{4} = \dfrac{y}{5}$，把 $\dfrac{x}{4} = \dfrac{y}{5}$ 代入到 $\dfrac{x}{4} - \dfrac{y}{10} = \dfrac{2}{3}$，可得 $y = \dfrac{20}{3}$，所以 $x = \dfrac{16}{3}$，则原方程组的解是 $\begin{cases}x = \dfrac{16}{3},\\y = \dfrac{20}{3}.\end{cases}$