If a temperature scale were bassed on the freezing point (5.5 degrees Celsius) and boiling point (80.1 degrees Celsius) of benzene, and the temperature difference between these two points was divided into 50 units called degrees X, what would be the freezing and boiling points of water in degrees X?
If I understand your question correctly, you are trying to create a new temperature scale based on the freezing and boiling points of Benzene.
Let's look at the corresponding points on each scale (X and Celsius)
Freezing Point: 0 X = 5.5 C
Boiling Point: 50 X = 80.1 C
The range in X degrees is 50, but in Celsius it is 74.6 degrees (=80.1-5.5)
That means that the X degrees are larger than the Celsius degrees, since there are fewer of them to cover the same temperature range. 1 X degree = 74.6/50 = 1.492 C degrees.
So now you can create a linear relationship between the two scales.
Recall from algebera, the slope of a line, m, can be expressed m = (Y1-Y2)/(X1-X2)
What that means is that the ratio of the two degree ranges from the freezing point to the boiling point of benzene is the same as the slope of the line relating the two scales.
Let's replace Y with C, to represent degrees C, and let the X's represent degrees X. We know the freezing and boiling points lie on the line, so lets use them for Y1, Y2, X1, X2:
m = (80.1-5.5)/(50-0) = 1.492, which agrees with what we found above.
For other points, we are going to replace one Y and one X with a known value and an unknown.
For example, say we want to know how many degrees X is 20 C.
We could use either of these equations.
1.492 = (80.1- 20)/ (50 - X)
1.492 = (20- 5.5)/ (X - 0)
They both return the same answer when solved for X, but both are a little awkward (20 C = 9.718 X) So, we take one of them and rearrange it to be nice and easy. The bottom one is the easiest to work with. It becomes:
Temp (X) = (Temp C - 5.5) / 1.492
I will leave it to you to rearrange for the conversion of X to C.
I hope this helps.