In order to model the complete two-compartment system, we must be able to solve equation (4). In section 2.2, we solved the equation by multiplying by two different weights and then dividing, and we can take a similar approach with the full two-compartment equation. We can weight equation (4) with three different weights and then integrate:
By multiplying equation (7) by and equation (9) by , and then subtracting the two, we may eliminate the V0 term. A similar operation can be performed on equation (8) and equation (9). This leaves two equations that do not contain V0. They may then be divided to produce:
The K1 term cancels out of both the numerator and denominator of equation (10), leaving an equation that only involves k2. As with the equation in section 2.2, this is very difficult to solve for k2. Therefore, a look-up table was again used. Once the table matching values of k2 with values of the right hand side of equation (10) has been created, we may evaluate k2 through simple lookup. With the k2 data computed, finding K1 is simply a matter of evaluating either the numerator or denominator of equation (10) without cancelling K1. With both K1 and k2 known, we may find V0 by evaluating equation (4).