Answer:
[tex]S=\dfrac{-2r(q+3r)}{q+2r}[/tex]
Step-by-step explanation:
You are given two equalities [tex]S=p-3r[/tex] and [tex]\dfrac{1}{p}-\dfrac{2}{q}=\dfrac{1}{r}[/tex]
Express p from the second equality:
[tex]\dfrac{1}{p}-\dfrac{2}{q}=\dfrac{1}{r}\\ \\\dfrac{1}{p}=\dfrac{1}{r}+\dfrac{2}{q}\\ \\\dfrac{1}{p}=\dfrac{q+2r}{rq}\\ \\\dfrac{p}{1}=p=\dfrac{rq}{q+2r}[/tex]
Substitute it into the first equality:
[tex]S=\dfrac{rq}{q+2r}-3r=\dfrac{rq-3r(q+2r)}{q+2r}=\dfrac{rq-3rq-6r^2}{q+2r}=\dfrac{-2rq-6r^2}{q+2r}=\dfrac{-2r(q+3r)}{q+2r}[/tex]
