Question

I need to solve for C the formula for temperature scale conversion

Answer 1

We need to rewrite the equation

$F=\frac{9C}{5}+32$

isolating C on the left side of the equation.

First, we can subtract 9C/5 from both sides of the equation. We obtain:

$\begin{gathered} F-\frac{9C}{5}=\frac{9C}{5}+32-\frac{9C}{5} \\ \\ F-\frac{9C}{5}=32 \end{gathered}$

Then, we can subtract F from both sides of the equation. We obtain:

$\begin{gathered} F-\frac{9C}{5}-F=32-F \\ \\ -\frac{9C}{5}=32-F \end{gathered}$

Then, we can multiply both sides by -5/9. We obtain:

$\begin{gathered} -\frac{9C}{5}\cdot(-\frac{5}{9})=(32-F)\cdot(-\frac{5}{9}) \\ \\ (-1)(-1)\frac{9}{9}\cdot\frac{5}{5}C=(32-F)\cdot(-\frac{5}{9}) \\ \\ C=(32-F)\cdot(-\frac{5}{9}) \end{gathered}$

Then, we can distribute the factor -5/9 over the sum 32-F on the right side:

$\begin{gathered} C=32\cdot(-\frac{5}{9}_{})-F\cdot(-\frac{5}{9}) \\ \\ C=-\frac{160}{9}+\frac{5}{9}F \end{gathered}$

Therefore, the answer can be written as

$C=-\frac{160}{9}+\frac{5}{9}F$

Or as

$C=\frac{5}{9}(F-32)$