I am returning to college after many years...so please forgive my stupidity. I have a problem that I must explain why it is unsolvable. Trouble is I think it is solvable. Help is greatly appreciated!

Here it is:
The sum of two numbers is 87. What are the numbers?

While I am also rusty, I think I can explain.
I suspect that you are thinking that the two numbers are both positive numbers. I also suspect that you are thinking that the two numbers are both integers and are not fractions nor do they contain fractions. If the problem gave that as one of the conditions, you would be right. That problem can be solved. The set of solutions would be (0+87) and (1+86) and (2+85) and on to (43+44). (Anything after that would simply be repetitious. 1+86 is the same as 86+1.)

But the problem did not limt the solution to non-fractions, nor did it limit the solutions to only positive integers. Since fractions can be used, you have an infinite number of solutions.
Ex. 1 1/2 + 85 1/2, 1 1/31 + 85 30/31, 1 501/502 + 85 1/502, ...
More examples:
86 1/2 + 1/2 = 87
86 1/3 + 2/3 = 87
86 1/4 + 3/4 = 87
86 1/5 + 4/5 = 87

And on and on, not only for 86, but every number less than 86. This series can go on endlessly, meaning that for every integer you choose, you can add an infinite number of fractions.

Even if the problem did disallow fractions, there are still an infinite number of solutions.
Using negative numbers (those less than zero), you can see that 100 plus a negative 13 equals 87.
100 + (-13) = 87
So for every positive number, there is a negative number that, when added, will produce 87 as the sum.
Examples:
1,000,000 + (-999,913) = 87
1,000,001 + (-999,914) = 87

And on and on. So this way also means that there are an infinite number of solutions.

In both cases (using fractions or using negative numbers), there exists an infinite number of answers. (This next part I am shaky about. I am sure that someone with more knowledge of Mathematics than I will correct me if I am wrong.) Since there are an infinite number of solutions, the equation is regarded as unsolvable.

Good luck at school. I went back at 43 with 8 semester hours credit*. At that time, I was a single parent, raising an 8 year old. Three years and three months later, I had my Bachelor's. Yes, it was a lot of work (and the work also included my part-time job), yes, it meant long nights, and yes, it was worth it, if for no other reason than my son got to see Daddy studying, staying up late to finish a papers, and graduate. My son is now a senior in college, majoring in something he loves. It was worth every minute.

Oh, by the way - Welcome to AnswerPool! I hope it fills your needs while you are at school, and I hope it also provides some relaxation as well. If you have any questions, please email Admin@AnswerPool.com.

*I spent 3 years getting those 8 hours credit. What can I say? I was young, and it was the 60s.

Welcome to AP, mommalisa. To add a postscript to DG's explanation. I would point out tha the problem as stated involves two unknowns. My first thought about that situation is that in order to solve for two unknowns you must have two equations. My history is even more ancient and fractured than DG's, so I wont swear to my rule, but I think it is correct. The problem, restated, could have been: the sum of 2 numbers is 87 and the difference between them is 5. Calling them x and y, write x+y=87 and x-y=5. Adding the two equations gives 2x=92, which solves as x=46. Then it becomes obvious that 87-46=41, so the two numbers are 46 and 41 and the solution can be readily proved. Q.E.D. Good luck with your education.