Can you solve this different kind of algebra problem?

The two heaviest together are 71 pound, the two lightest together are 58. From this we can find the middle one is 33.

Let’s try 36 for the heaviest (at least one must weigh more than 35).

71–36 = 35

so one must be 35

But then the 70 pair is problematic…

so let’s try 37.

71–37 = 34
70–37 = 33
So the middle one is right.

Now for the lower 2….

30, and 28….
Can’t get 60.
31 and 27

27, 31, 33, 34, 37

I believe that’s it.

If we call the weights of the five children in ascending order a,b,c,d and e
Then it follows that…

a+b=58
a+c=60
a+d=61
a+e=64
b+c=64
b+d=65
b+e=67
c+d=68
c+e=70
d+e=71

Adding both sides together 4(a+b+c+d+e) =648
As we know from above what any four of these weights add up to we can then calculate the 5th

The weights of the children are therefore 27,31,33,35 and 36Kg

@flutherother , How did you determine b+c and a+e. How do you know that a+e is less than c+d?

Order the weights a<=b<=c<=d<=e.
If you think about it, you will see that the smallest sum is a+b and the next smallest is a+c. Similarly the largest sum is d+e and the next largest is c+e. That is the only immediate conclusion we can come to.

a+b = 58
a+c=60
c+e=70
d+e=71

That is four equations for five variables. We need one more equation. I did not see @flutherother ‘s trick of using the sum of all the pairs of weight sums. I used a much more convoluted approach. Getting 4(a+b+c+c+e)=648 provides the extra equation.
a+b+c+d+e = 162.
We then have c =(a+b+c+d+e) – (a+b) – (d+e) = 162 – 129 = 33. Having c, it is easy to use the equations to find the remaining values.

I assumed too much and got the wrong answer. @6rant6 got the right answer and @LostInParadise has the right method.