Given 5 equations, can I solve for 6 unknowns? Background
I am trying to extract data from scientific publications. Sometimes an experiment can test two factors, e.g. a standard two-way factorial ANOVA. 
Call these factors $A$ and $B$. If factor $A$ has two levels and $B$ has three, there are six total treatments.
If the effects of $A$ and $B$ are significant, but there is no interaction, only the 'Main' effects might be presented, e.g. five results, one for each level of $A$ and one for each level of $B$, averaged across all of the levels of the other factor.
Here is an example from Ma 2001 Table 2, in which $A$ would be the row spacing and $B$ would be the nitrogen rate. 

Thus,
$$7577 = \frac{X_{A_{20},B_{0}} + X_{A_{20},B_{112}} + X_{A_{20},B_{224}}} {3}$$
$$9186 = \frac{X_{A_{80},B_{0}} + X_{A_{80},B_{112}} + X_{A_{80},B_{224}}} {3}$$
$$3706 = \frac{X_{A_{20},B_{0}} + X_{A_{80},B_{0}}} {2}$$
$$9402 = \frac{X_{A_{20},B_{112}} + X_{A_{80},B_{112}}} {2}$$
$$12038 = \frac{X_{A_{20},B_{224}} + X_{A_{80},B_{224}}} {2}$$
Question
Is it possible to calculate the means of each of the six treatments $X_{A,B}$, for $A\in[20,80]$ by $B\in[0,112,224]$ from these results? 
 A: You can do it if you make some assumption to reduce the number of unknowns to five.  You are saying you have an array
$$\begin{array} {ccc}  & 20 & 80 \\ 0 & a & b \\ 112 & c & d \\ 224 & e & f \end{array}$$ 
where $a$ through $f$ are what you want to solve for. If the effects are independent and additive, you would expect $b-a=d-c=f-e$, $e-c=f-d$, and $c-a=d-b$.  These reduce the data to only three values, which you can check for consistency.  But without at least one more relation, you will get a one-dimensional continuum of solutions.
It sounds like you believe 3706 is some sort of weighted average of $a$ and $b$ and similarly for the other entries.  Is that right?
A: The general rule is that $n$ equations allows you to solve for $n$ unknowns. So I don't think you'll be able to recover each of the original 6 data points. The best you can do is produce a set of constraints that, given any one of the 6 data points (or a relationship between them that isn't redundant with what you already have), would allow you to find the remaining 5.
A: [Update Note] I saw the related question at mathoverflow. There it seemed we were dealing with a frequencies-table, so I repeated that scheme here. But now I see the question is focused on means in a two-factor anova. I'll see, whether that two concepts can be interchanged here; for instance a 1:1-reference should only be possible if the coefficients under treatments (means?) are based on the same number of observations. Possibly it is better to delete this answer later [end note]
Here is a solution. I computed the "expected frequencies" based on your values, where I compensated the * .../3* and the * .../2*-divisions. Also I corrected 9186 to 9817 to make the totals consistent.
$ \begin{array} {rrrrrrr}
  & & B:  '0' &    '112' & '224'   &| &      (all) & \\\
---    &+&----&-----&-----&+&---& &\\\
A:'20' &|& 3350.08&  8499.04 & 10881.88 &|& 22731  &/3 = 7577 \\\   
'80'   &|& 4061.92& 10304.96 & 13194.12 &|& 27561 & /3 = 9187 \\\    
---    &+&----&-----&-----&+&--- & \\\     
(all)   & &  7412& 18804 & 24076 &|& 50292  & \\\ 
        & &  /2=3706&   /2=9402  &  /2=12038     & &      & \\\
 \end{array}
$ 
