## Elementary Algebra

$F=59$
We are trying to isolate $F$ in the equation $C=\frac{5}{9}(F-32)$. To do so, we must first multiply by $\frac{9}{5}$, the reciprocal of $\frac{5}{9}$, on each side: $$\frac{9}{5}C=\frac{9}{5}\times\frac{5}{9}(F-32)$$ $$\frac{9}{5}C=F-32$$ Next, we must add $32$ to each side of the equation: $$\frac{9}{5}C+32=F-32+32$$ $$\frac{9}{5}C+32=F$$ Finally, we plug in the given value for $C$ into the equation and follow the order of operations to simplify for $F$: $$F=\frac{9}{5}(15)+32$$ $$F=27+32$$ $$F=59$$