Taking random values will not do
Algebra
Level
3
\[\large{\begin{align*}x+\dfrac{1}{yz} &= \dfrac{1}{5}\\ y+\dfrac{1}{xz} &= -\dfrac{1}{15}\\ z+\dfrac{1}{xy} & =\dfrac{1}{3} \end{align*}}\] If \(x,y\) and \(z\) are real numbers which satisfy the given equations above, find \(\dfrac{z-y}{z-x}\).
by
Abhay Kumar