F=9/5C+32, solve for C
\(\dfrac{9}{5}C+32=F\ \ \ \ |subtract\ 32\ from\ both\ sides\\\\\dfrac{9}{5}C=F-32\ \ \ \ |multiply\ both\ sides\ by\ \dfrac{5}{9}\\\\\boxed{C=\frac{5}{9}(F-32)}\)
F=9/5C+32F-32=9/5C9/5C=F-329C=5(F-32)C=\(\frac{5(F-32)}{9}\)
