Pairwise Puzzler

The five numbers can be calculated simply by using simultaneous equations.

Let's call the lowest number a, the second lowest b etc.. and we'll place the sums from smallest to largest in series s.
We know straight away that a + b = s1, as the smallest sum must be made up of the smallest pair.
Also the largest sum is clear: d + e = s10
We can also work out that a + c = s2, as this is the second smallest sum; a and c are the smallest two numbers which haven't yet been paired.
Similarly, c + e = s9 (the second largest).

So we now have five unknowns and four equations... we just need one more equation.
If we look at the sum of all the sums in terms of the five original numbers ((a + b) + (a + c)...(e + f)),
this simplifies to 4 (a + b + c + d + e).
So therefore a + b + c + d + e = sum (s) / 4
Here we have the final equation, and we can now calculate the five numbers through simple algebra.

So let's make each letter in turn the subject of an equation.

We can do this easily for c, using substitution: a + b = s1, d + e = s10, and a + b + c + d + e + f = sum (s) / 4;
therefore s1 + s10 + c = sum (s) / 4
So c = sum (s) / 4 - s1 - s10

Now that we have c, all of the other letters can be expressed in terms of c:
a = s2 - c
b = s1 - (s2 - c) = s1 - s2 + c
e = s9 - c
d = s10 - (s9 - c) = s10 - s9 + c

These equations can all be written, by substituting c, in terms of the sums:
a = s1 + s2 + s10 - sum (s) / 4
b = sum (s) / 4 - s2 - s10
c = sum (s) / 4 - s1 - s10
d = sum (s) / 4 - s1 - s9
e = s1 + s9 + s10 - sum (s) / 4

So to find the original five numbers given only the sum of each pair, just arrange the sums from smallest to largest and substitute them into the equations above.

The answer to the first question is 0,2,6,7,9!!!!!

Having tackled the 4- and 5-number versions of this problem, I couldn't see an obvious way to generate the 6th equation which would allow direct solution via simultaneous equations. Five equations are simple enough to find, as Mr Redman showed very clearly in his comment, but without the sixth (which I hope someone will enlighten me about at some point), the system would be under-determined. So, without an elegant solution to hand, I tried an inelegant one instead...

About 10 lines of matlab code allowed me to generate random collections of six integers (within a reasonable range to keep things somewhat manageable), generate all 15 of their pairwise combinations and sums, and test the resulting sums against the given sums. About 25 seconds of computer time gave me the following answer (which I have checked by hand just to be sure): 2 5 8 11 13 19

I know that this technique will scale very poorly as the size of problem increases - I certainly wouldn't want to attempt it for pairwise combinations of 10 numbers, for example - and would welcome any suggestions regarding a more subtle approach, but feel that a computational approach does have a place even on this website!