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# Converting Fahrenheit to Celsius (or vice versa)

Q:  How can I make up an equation that will convert Fahrenheit to Celsius (or vice versa)??

A:  Converting Fahrenheit to Celsius is a linear process.  What does that mean?  The equation is a “y = mx + b” type… So,

C = mF + b

(C = Celsius and F = Fahrenheit)…. I could also do F = mC + b… But, I am going to stick with the first method I picked.

Now, I need a few data points… Things that I know about Fahrenheit and Celsius…. I know that

32 Fahrenheit = 0 Celsius [freezing point]

and that

212 Fahrenheit = 100 Celsius [boiling point]

So, I have the data points (32, 0) and (212, 100) to use in my equation:

C = mF + b

How do I calculate the slope (m)?  Use our regular slope equation:

m = (y2 – y1) / (x2 – x1)
But, in our case, we are using “F” and “C”… So,

m = (C2 – C1) / (F2 – F1)

m = (100 – 0) / (212 – 32)

m = 100 / 180

m = 5 / 9

So, we know that:

C = mF + b

C = (5/9)F + b

Now, plug in a point to solve for b.  I will use the point (32, 0):

0 = (5/9)(32) + b

0 = 160/9 + b

-160/9 = b

So,

C = (5/9)F – 160/9

For example, let’s say you were given 15 degrees Fahrenheit and asked to convert it to Celsius: Simply plug in 15 for F:

C = (5/9)(15) – 160/9  and solve!

OR, let’s say you were given 20 degrees Celsius and asked to convert it to Fahrenheit: Simply plug in 20 for C:

20 = (5/9)F – 160/9  and solve!