Converting between Celsius and Fahrenheit

Date: 06/23/2009 at 15:47:27
From: Jeff
Subject: Why does the formula °F = 9/5(°C + 40) - 40  work

The formulas °F = 9/5(°C + 40) - 40 and °C = 5/9(°F + 40) - 40 both 
work and are easy to remember because all you do is change 5/9 to 9/5 
depending on which way you want to convert and 9/5 > 5/9 and °F > °C. 

With the regular equations I can never remember do I add then multiply
or multiply then add or is it subtract?  So the above equations remove
the confusion.

But when I tried to derive these formulas with algebra using F = mC +
b or C = mF + b, I just keep getting the standard forms of °F = 9/5°C
+ 32 and °C = 5/9(°F - 32).

Is there some other way I should be approaching this?

Date: 06/23/2009 at 21:39:24
From: Doctor Rick
Subject: Re: Why does the formula °F = 9/5(°C + 40) - 40  work

Hi, Jeff.

The formula F = (9/5)C + 32 is in slope-intercept form, so that's what
you get when you work with the slope-intercept form y = mx + b.

You can use another form for the equation of a straight line--the 
point-slope form:

  y = m(x - h) + k

where m is the slope and (h, k) is a point on the line.  One point 
that we know is on the line is (-40, -40)--that is, -40 F and -40 C 
are the same temperature.  Knowing that 9/5 degree F = 1 degree C, 
the slope is 9/5, so plugging these values into the form, we get:

  F = (9/5)(C - -40) + -40
  F = (9/5)(C + 40) - 40

Likewise, if C is the independent variable, the slope is 5/9.  The 
known point is the same, (-40, -40), since switching coordinates 
doesn't change it.  Thus we get

  C = (5/9)(F - -40) + -40
  C = (5/9)(F + 40) - 40

- Doctor Rick, The Math Forum
  http://mathforum.org/dr.math/ 

Date: 06/24/2009 at 16:17:37
From: Jeff
Subject: Thank you (Why does the formula °F = 9/5(°C + 40) - 40  work)

Thank you very much for your solution to my problem.  I hope that you
can see why I prefer this form of the equation (less to remember =
easier to remember).