An alternative approach to the accurate conversions, that may be easier
to remember because it is more symmetric, is to first shift the
temperatures so that the common value (-40) is the zero point, then
scale, then shift back, that is,
\begin{align*}
f & = (c + 40) \cdot \frac{9}{5} - 40, \\
c & = (f + 40) \cdot \frac{5}{9} - 40.
\end{align*}
The formula "looks" the same for each direction, just with inverted
scaling factors.

## Geoff Bailey said:

An alternative approach to the accurate conversions, that may be easier to remember because it is more symmetric, is to first shift the temperatures so that the common value (-40) is the zero point, then scale, then shift back, that is, \begin{align*} f & = (c + 40) \cdot \frac{9}{5} - 40, \\ c & = (f + 40) \cdot \frac{5}{9} - 40. \end{align*} The formula "looks" the same for each direction, just with inverted scaling factors.