解:将$\frac {2x + 3y}{2}、$$\frac {3x + 2y}{5}$分别看作一个整体,
分别设为$u,$$v,$得$\begin {cases}u = v + 2\\3u = 2v + 6\end {cases},$
将$u = v + 2$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
$x=\frac {4-3×\frac {12}{5}}{2}=\frac {20 - 36}{10}=-\frac {8}{5}。$
所以原方程组的解为$\begin {cases}x = -\dfrac {8}{5}\\y = \dfrac {12}{5}\end {cases}$