[tex]\dfrac{1}{p}+\dfrac{1}{q}=\dfrac{1}{f}\qquad|\text{subtract}\ \dfrac{1}{q}\ \text{from both sides}\\\\\dfrac{1}{p}=\dfrac{1}{f}-\dfrac{1}{q}\\\\\dfrac{1}{p}=\dfrac{(1)(q)}{(f)(q)}-\dfrac{(1)(f)}{(f)(q)}\\\\\dfrac{1}{p}=\dfrac{q-f}{fq}\Rightarrow\dfrac{p}{1}=\dfrac{fq}{q-f}\\\\\boxed{p=\dfrac{fq}{q-f}}[/tex]
