Elementary Algebra

Published by Cengage Learning
ISBN 10: 1285194055
ISBN 13: 978-1-28519-405-9

Chapter 3 - Equations and Problem Solving - 3.2 - Equations and Problem Solving - Problem Set 3.2 - Page 106: 41



Work Step by Step

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 $C$: $$C=\frac{5}{9}(68-32)$$ $$C=\frac{5}{9}(36)$$ $$C=20$$
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