Taking random values will not do

Algebra Level 3

$\large{\begin{aligned}x+\dfrac{1}{yz} &= \dfrac{1}{5}\\ y+\dfrac{1}{xz} &= -\dfrac{1}{15}\\ z+\dfrac{1}{xy} & =\dfrac{1}{3} \end{aligned}}$ If $x,y$ and $z$ are real numbers which satisfy the given equations above, find $\dfrac{z-y}{z-x}$ .