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Please avoid the trivial $x=y=z=a=b=5$. Try looking for a solution where $x≠y≠z≠a≠b$ or if not, look for one where one variable equals to another, but explain your reasoning. The girl was covering "unit fractions" in her class.

Was this a requirement in the homework problem? The easiest way to prove that there is more than one solution to a problem is to find more than one solution to the problem, even if the additional solution is trivial. A starting point of any arbitrary number of unit fractions that sum to $1$ and that also can be combined into other unit fractions is a very valid starting point.

Case in point: I would propose that this be solved at the fifth-grade level by splitting fractions, with a little special knowledge to produce unique values. $1 = 1/2 + 1/2$. Now we need to come up with three more fractions and differentiate the two we have. We do this by continuing to split, with some clever knowledge of prime factors.

For instance, if a number in the denominator is divisible by both $2$ and $3$, then you can combine two or three of that unit fraction, and when reducing to lowest terms you will get a new unit fraction. $6$ is the simplest example; $1/6 + 1/6 = 1/3$ and $1/6 + 1/6 + 1/6 = 1/2$. This is useful to us: $1/6 + 1/3 = 1/2$, so $1/6 + 1/3 + 1/2 = 1$. Now we just need two more unique fractions. Well, let's split up the largest fraction, $1/2$, into more pieces than we tried before. If we divide it into 12 units, each of those will be $1/24$ of the whole. $2/24 = 1/12, 3/24 = 1/8$, and $6/24 = 1/4$. We can also combine $4$ and $8$ of these, but those two produce the fractions $1/6$ and $1/3$ which we already have. Now, by serendipity (or not), $1 + 2 + 3 + 6 = 12$. That would give us $1/24 + 1/12 + 1/8 + 1/4 = 1/2$. That's four of the five we need, and $1/2$ (the remaining fraction of the whole) isn't spoken for yet and can be produced by recombining $1/3$ and $1/6$, so $1/2 + 1/4 + 1/8 + 1/12 + 1/24 = 1$, and thus $\{2,4,8,12,24\}$ is a valid solution.

This generalizes into the following statement: find five numbers, $j,k,m,n$, and $p$, that are all factors of an arbitrary $z$, and that sum to $z$. Then $1/x=j/z$, $1/y=k/z$, and so on.

A good z-value to try in this case is $100$, which turns the fractions into simple integer percentages. $100$ has the following prime factorization: $2^2*5^2$. Each unique combination of those factors, plus the universal factor $1$, is a factor of $100$ and thus a possible value for the set of $5$ we need. There are 8 factors of $100$, not including $100$; $1, 2, 4, 5, 10, 20, 25$, and $50$. We can now apply a variation of the value-splitting; find numbers in this set that will sum to other numbers in the set. $100=50+50$. $50=25+25$. $25=20+5$, and $5=4+1$. Putting these identities together, $1+4+20+25+50 = 100$. Dividing everything by $100$ and finding lowest terms gives $1/100+1/25+1/5+1/4+1/2 = 1$, so $\{2,4,5,25,100\}$ is another valid solution.

Pretty much any number with $4$ or more factors other than $1$ and itself is a candidate z-value. It helps to have a few extra factors laying around, as was shown from the example above; a number with only $4$ unique factors, say $20$ $(2, 4, 5, 10)$, is unlikely to have those factors all sum to the original number as well (in this case the sum of those four factors is $21$; close but no cigar). It also helps if the prime factorization of the number (a graspable concept for a fifth grader, with multiplication, division and primes under her belt) includes more than one prime factor; this allows you to do the "divide by one, then multiply by the other" trick to produce unique denominators.

The are multiple, but not infinite, total solutions to this problem; the numbers involved are naturally bound by the two conditions; that natural numbers are used, and that only reciprocals of those natural numbers are allowed (no numerators other than 1). Other answers have empirically found all possible answers.