解：將?$\frac {2x + 3y}{2}、$??$\frac {3x + 2y}{5}$?分別看作一個(gè)整體，
分別設(shè)為?$u，$??$v，$?得?$\begin {cases}u = v + 2\\3u = 2v + 6\end {cases}，$?
將?$u = v + 2$?代入?$3u = 2v + 6，$?得
?$3(v + 2)=2v + 6，$?
解得?$v = 0，$?
把?$v = 0$?代入?$u = v + 2，$?得?$u = 2。$?
即?$\begin {cases}\frac {2x + 3y}{2}=2\\\frac {3x + 2y}{5}=0\end {cases}，$?整理得?$\begin {cases}2x + 3y = 4\\3x + 2y = 0\end {cases}，$?
由?$2x + 3y = 4$?得?$x=\frac {4 - 3y}{2}，$?
代入?$3x + 2y = 0$?得?$3×\frac {4 - 3y}{2}+2y = 0，$?
解得?$y = \frac {12}{5}，$?
把?$y = \frac {12}{5}$?代入?$x=\frac {4 - 3y}{2}$?得
?$x=\frac {4-3×\frac {12}{5}}{2}=\frac {20 - 36}{10}=-\frac {8}{5}。$?
所以原方程組的解為?$\begin {cases}x = -\dfrac {8}{5}\\y = \dfrac {12}{5}\end {cases}$?
