## Elementary Algebra

$C=20$
We are trying to isolate $C$ in the equation $F=\frac{9}{5}C+32$. To do so, we must first subtract $32$ from each side because $32$ is added to the $C$ term: $$F-32=\frac{9}{5}C+32-32$$ $$F-32=\frac{9}{5}C$$ Next, we must multiply by $\frac{5}{9}$, the reciprocal of $\frac{9}{5}$, on both sides of the equation: $$\frac{5}{9}(F-32)=\frac{5}{9}(\frac{9}{5}C)$$ $$\frac{5}{9}(F-32)=C$$ Finally, we plug in the given value for $F$ into the equation and follow the order of operations to simplify for $C$: $$C=\frac{5}{9}(68-32)$$ $$C=\frac{5}{9}(36)$$ $$C=20$$