SAT Question 133f3e41: Nonlinear Equations In One Variable And Systems Of Equations In Two Variables answer and explanation with Desmos solution

Choice C is correct. Multiplying each side of the given equation by $\frac{1}{20}$ yields $\frac{1}{20}\left(\frac{20}{p}\right)=\frac{1}{20}\left(\frac{20}{q}-\frac{20}{r}-\frac{20}{s}\right)$. Distributing $\frac{1}{20}$ on each side of this equation yields $\frac{20}{20p}=\frac{20}{20q}-\frac{20}{20r}-\frac{20}{20s}$, or $\frac{1}{p}=\frac{1}{q}-\frac{1}{r}-\frac{1}{s}$. Adding $\frac{1}{r}+\frac{1}{s}$ to each side of this equation yields $\frac{1}{s}+\frac{1}{r}+\frac{1}{p}=\frac{1}{q}$. Multiplying $\frac{1}{s}$ by $\frac{pr}{pr}$, $\frac{1}{r}$ by $\frac{ps}{ps}$, and $\frac{1}{p}$ by $\frac{rs}{rs}$ yields $\frac{pr}{prs}+\frac{ps}{prs}+\frac{rs}{prs}=\frac{1}{q}$, which is equivalent to $\frac{pr+ps+rs}{prs}=\frac{1}{q}$. Since $\frac{pr+ps+rs}{prs}=\frac{1}{q}$, and it's given that $p$, $q$, $r$, and $s$ are positive, it follows that the reciprocals of each side of this equation are also equal. Thus, $\frac{prs}{pr+ps+rs}=\frac{q}{1}$, or $\frac{prs}{pr+ps+rs}=q$. Therefore, $\frac{prs}{pr+ps+rs}$ is equivalent to $q$.