How to solve temperature readings in a faulty thermometers using correct thermometer readings? Here is the context. I have a infrared thermometer and a digital thermometer. My daughter (2 years old) is afraid of anything that looks like a syringe, for obvious reasons. (Please don't judge my parenting skills, though :)) So, we can only use the infrared thermometer most of the time and it doesn't give the correct temperature. So, I tried to find a formula to be able to be able to match the readings with the digital (correct) one. What I did was, took two different person temperature readings (all readings are in F) and tried solving the quadratic equations. For 98.3 and 99 degrees the faulty one gives 99.1 and 100.4 degrees, respectively. So, the equations I derived are:
98.3X - 99.1Y  = 0 &
99X   - 100.4Y = 0
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0.7X  - 1.3Y   = 0

X = (1.3/0.7)Y

Substituting the value of X in one of the equation:
99(1.3/0.7)Y - 100.4Y = 0
Y = 1/83.457 = 0.011982

Substituting value of Y, we get value of X:
99X - 100.4*0.011982 = 0
X = 0.012152

Now we should be able to use these X & Y values to get the correct reading for a wrong reading.
if Tw = 97.3 (w stands for wrong) then correct reading Tc should be:
Tc = Tw*(Y/X)
   = 97.3*0.986010533
   = 95.93

But the actual reading is 96.9.
Have I got this completely wrong or am I missing something?

Let the faulty temperature be $Y$ and the correct temperature be $X$, then assuming a linear model, we have
$ Y = a X + b $
Substituting the two pairs of temperatures that we have
$ 99.1 = a (98.3) + b $
$ 100.4 = a (99) + b $
Which solves to $a = \dfrac{13}{7} , b = - \dfrac{2921}{35} $
From which
$X = \dfrac{1}{a} (Y - b) = \dfrac{7}{13} (Y + \dfrac{2921}{35} )$
Plug in $Y = 97.3$, you get
$X = \dfrac{7}{13} ( 97.3 + \dfrac{2921}{35} ) = 97.33 $
Since the actual temperature measured is $96.9$, it means our linear model is not $100\%$ correct, and also that we need more decimals in the our readings of the actual correct temperature and the faulty one.  However, note that the percent difference between the calculated temperature and the measured one is
$ \dfrac{ 97.33 - 96.9 }{96.9} \times 100 \% = 0.44 \% $
which is a very good percent error.
