Hint: Sum up all pairs and see what happens!

Explanation:

Let's say our numbers are a, b, c, d, e.

Then our pairs will be: a + b, a + c, a + d, ..., d + e.

If we sum up all the pairs we've 4 * (a + b + c + d + e).

If we divide by 4, then we have x = a + b + c + d + e.

And from now if we want to get, for example e, we choose a + b and c + d and x - (a + b) - (c + d) = e.

And you can do that for all the five numbers!

Hope it helps! Have nice day :)

I am not a mathematician but i want to share a process that i think it may work for You

lets do some math . As you mention 5 numbers are X1 , X2 , X3 , X4 , X5 and what we have ( X1 + X2 ) , ( X1 + X3 ) + (X1 + X4 ) + ( X1 + X5 ) + ( X2 + X3 ) + ( X2 + X4 ) + ( X2 + X5 ) + ( X3 + X4 ) + ( X3 + X5 ) + ( X4 + X5 ) == TOTAL_OF_ALL_PAIR_NUMBERS -> 4X1 + 4X2 + 4X3 + 4X4 + 4X5 == TOTAL_OF_ALL_PAIR_NUMBERS

SO ( X1 + X2 + X3 + X4 + X5 ) = ( TOTAL_OF_ALL_PAIR_NUMERS)/4 == TOTAL

`now we have total of all numbers . Now take any pair of number . Assume its ( X1 + X2 ) . So what we have now X3 + X4 + X5 = total — ( X1 + X2 ) . As we take one pair there other nine pair remaining . take 3 pair and check if these 3 pairs can only form by X3 + X4 + X5 or not . How we can ensure that this 3 pair can only form by these X3 + X4 + X5 . If and only if ( X3 + X4 ) + ( X3 + X5 ) + ( X4 + X5 ) == 2 * ( X3 + X4 + X5 ) . Thats means TOTAL — ( X1 + X2 ) == ( X3 + X4 + X5 )/2 . When we sure about our chosen 3 pair we can easily find what is the value of X3 , X4 and X5 are . As X3 = ( Total — ( X1 + X2 ) + ( X4 + X5 ) . Do the process one more time . Surely we find one unknown number X1 or X2 ' . There should be more simple process .