Let $x$, $y$, and $z$ be real numbers such that $x+y+z=6$ and $\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=2$. Find the value of $\dfrac{(x+y)}{z}+\dfrac{(y+z)}{x}+\dfrac{(x+z)}{y}$.