$解：∵y_{1}与x+1成正比例，∴设y_{1}=k_{1}(x+1)$
$∵y_{2}与x-1成正比例，∴设y_{2}=k_{2}(x-1)$
$∵y=y_{1}+y_{2}，∴y=k_{1}(x+1)+k_{2}(x-1)$
$∵当x=2时，y=9；当x=3时，y=14$
$∴\begin{cases}{3k_{1}+k_{2}=9}\\{4k_{1}+2k_{2}=14}\end{cases}，解得\begin{cases}{k_{1}=2}\\{k_{2}=3}\end{cases}$
$∴y与x的函数表达式为y=2(x+1)+3(x-1)，即y=5x-1$
