Answer
$x=-\dfrac{yz}{z-y}$
Work Step by Step
Using the properties of equality, in terms of $
x
,$ the given equation, $
\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{z}
$ is equivalent to
\begin{array}{l}\require{cancel}
xyz\left( \dfrac{1}{x}+\dfrac{1}{y} \right) =\left( \dfrac{1}{z} \right) xyz
\\\\
yz(1)+xz(1)=1(xy)
\\\\
yz+xz=xy
\\\\
xz-xy=-yz
\\\\
x(z-y)=-yz
\\\\
x=-\dfrac{yz}{z-y}
.\end{array}
