Do absorbances of different compounds at one wavelength sum up?
[&hellip;] I can see no way to use this equation, since i'm only given a wavelength and a whole bunch of coefficients.


We have all the data that we need.

You have already mentioned the Lambert-Beer law. 

For the solution of a single compound S in a cuvette of width d, the absorbance $E_\lambda$ at a particular wavelength $\lambda$ is proportional to the concentration $c$ of S.

\[E_\lambda = \log_{10}{\left(\frac{I_0}{I} \right)} = \epsilon_\lambda\cdot c \cdot d \]

A higher concentration of S will result in higher absorbance.

Now imagine that the light-absorbing molecules S are linked together by units that do not absorb the light at the wavelength observed.

Let's compare two equimolar solutions of 
$$\ce{-linker-S-linker-S-linker}$$ 
and
$$\ce{-linker-S-linker-S-linker-S-linker-S-linker-S-linker}$$

The second will show a higher absorbance, since the concentration of S is higher!

Now let's have a look at equimolar solutions of different compounds S1 and S2 and their absorbance.

We have outlined above that the absorbance $E_\lambda$ is proportional to the concentration of the light-absorbing species.

But apparently, different compound to not all absorb the same at a particular wavelength!

Exactly this is expressed with the molar absorption coefficient $\epsilon_\lambda$. The higher $\epsilon_\lambda$, the stronger the attenuation of incoming light. Note that $\epsilon_\lambda$ is 


always given for a particular wavelength
a compound-specific constant 


Let's tie the pieces together!


When light-absorbing units are linked together by 'blind' linkers, the effective concentration of the absorbing units counts.
Different compounds can absorb at the same wavelength, but due to their specific $\epsilon_\lambda$, they will do this to different amounts.


Consequently, we can sum up the combined absorbances in the following way:

$$E_\lambda =  \left(\epsilon_A \cdot c_A + \epsilon_C \cdot c_C + \epsilon_G \cdot c_G + \epsilon_T \cdot c_T \right)\cdot d$$