Table 2.

How to find x and y with matrix algebra.

The equation we need to solve is the following:
Number of people that tested positive in both groups (z) =
the number of true positives (a) + the number of false positives (c)
where
Number of people with an infection in disease group	x
Number of people without an infection in control group	y
Total number of people in the disease and control groups	x + y
Test sensitivity (se)	0.44
Test specificity (sp)	0.99
Number of true positives (a) = x * se	x * 0.44
Number of false positives (c) = y * (1 - sp)	y * (1-0.99)
Number of people that tested positive in both groups (z)	100 000
which means:
The equation we need to solve is	z = x * se + y * (1-sp)
with constraint	z =	100 000
	se =	0.44
	sp =	0.99
	y =	p*x
	p =	2.5
We can solve such equation in excel by using matrix algebra
The system of linear equations we should solve is	A * B = C
where
A =	0.44	0.01
	−2.5	1
B =	x
	y
C =	100 000
	0
A^-1 =	2.150538	−0.021505
	5.376344	0.946237
B = A ^ (-1) * C	x =	215 054
	y =	537 634
The relationship between y and x (p) = y / x	2.5
Number of people that test positive in the disease group = true positive (a) = x * se	94 624
Number of people that test positive in the control group = false positive (c) = y * (1-sp)	5 376
Number of people that tested positive in both groups (z) = a + c = x * se + y *(1-sp)	100 000
% of people that tested positive in both groups (zz) = (z / ( x + y) )*100	13.29
I have written a user defined function (udf) in VBA that does the above calculations automatically
Lyme1(se ; sp ; p ; z ; output) where output is either "x", "y" or "zz"
Lyme1(0.44;0.99;2.5;100000;"x")	x =	215 054
Lyme1(0.44;0.99;2.5;100000;"y")	y =	537 634
Lyme1(0.44;0.99;2.5;100000;"zz")	zz =	13.29