I’d say the algebra problem is buried under a story problem and encoded in annoying algebra story problem language.

I think before you start asking individual students to solve story problems that use expressions like “we have n > 1 people”, you ought to very clearly make sure understand that that is algebra-language for “there are 2 or more people”.

And I think the problem type is best presented in natural language and/or with models or drawings, not algebra.

Once you’ve done those things, I think then you could try to teach how algebraic language can be used to describe the problem and be a useful way to think about and write about such a problem. But I expect you will lose the understanding and/or attention of most of a typical American middle school class in the process. It may still be worth throwing at them, but I wouldn’t expect everyone to really get it.

I expect many middle schoolers will be trying to memorize (not understand) the problem pattern so they can mimic an answer on a test, and/or get frustrated and give up. At least, in the culture of American public middle school students I’ve had recent exposure to.

I agree with you about removing the algebra from the description of the problem.

The problem of memorizing and solving by rote is exactly what I am attempting to get around. The problem is not that difficult. Once you look at it in the right way, the actual algebra is fairly simple.

Maybe I would start out by asking to explain why the total number of crossings must be an odd number. This requires no algebra at all and should be apparent by solving for simple cases of n=2 and n=3. The final crossing is different from the others. Seeing why this is so provides a key insight into the solution of the problem

There is always the classic Mary is 2 year older than John. Their ages add up to 15. How old is John?

Mary needs 85 points to retire. Points are the sum of the persons age and years of service. If Mary started working when she was 23 at what age can she retire?

These can be solved without algebra but does give a couple of lessons in economics.
A bank offers 5% interest on savings. How many years will it take to double the savings?
or
A store charges 24% annual interest on money owed for a purchase. In how many years will the money would be double the original purchase price?

I like the second problem. It is a good example of how translating into algebraic notation gives an equation that can be solved for the variable.

This is one way of reasoning about the first problem that I gave: It may be much for middle school. Certainly a high school student be able to solve it.
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Boat crossings start out in groups of 2. Two people cross, one person gets off, and the other returns. That is a net change of one person.

At some point there will be two people remaining. It takes a single crossing to get them to the other side with no need to return. The total number of crossings is some number of pairs plus the final crossing and is therefore an odd number

Two people go across in the last crossing, leaving n – 2 to go across in pairs of crossings. There must be n -2 pairs of crossings, since each pair adds one person to the other side. The total number of crossings is therefore one last crossing plus n -2 pairs, or 1 + 2(n-2) = 2n – 3

The second problem is similar. This time 3 people go across on the last crossing. That leaves n – 3 people who go across in pairs of crossings. For each of these pairs of crossings, there is a net change of 2 people, since 3 people go across and one returns. Therefore there must be (n-3)/2 pairs of crossings. (n-3)/2 pairs is 2(n-3)/2 = n-3 crossings, to which we add the one last crossing, giving n – 3 +1 = n-2 crossings.

If it takes Bob 1 hour to mow the lawn, and it takes Blob 2 hours to mow the same lawn, how long will it take them if they mow the lawn together?