OK, every once in a while I go thru my old high school algebra and trig books to keep my mind exercised, but I’m not sure how this one is worked out:
Determine values for a and b such that the equation ax + b = 0 has the solution set 3/2.
Does this mean that a*3/2 + b = 0, and now find what values of a and b satisfy that? If that’s it, by my estimates, there would seem to be. . . a lot of answers.
Yes, and yes (assuming there isn’t some other context we’re missing).
It’s not clear to me from what you’ve quoted whether they want any one particular set of values for a and b or a description of all possible values. There would indeed be infinitely many pairs of values (all in the same proportion to one another).
Yes and yes. The solution set is “b such that b / a = -3/2” or “For a={1, 2, 3, …}, b={-3/2, -3, -9/2, …}
Stranger
In just a bit more more detail, a polynomial of the form ax + b = 0 describes a straight line. The technical term is “slope intercept format” which may jog some memory cells or give you some Google fodder.
We don’t know (yet) what angle that straight line makes on the standard “Cartesian” X-horizontal-Y-vertical graph. And we don’t know (yet) where that line crosses the Y axis. But we do know that the values of a & b determine those things and our job is to find those values.
Now the statement “The solution set is 3/2” means for any & every possible x input, the result is y = 3/2 = 1.5.
That’s probably enough to get you going. If not, bug us again. 
I’ll just mention that, in textbook math problems, it’s not uncommon to see one that has many possible right answers. This will often be noted in the problem itself, and if the textbook comes with an online question system, the system (if well-designed; I’ve seen some pretty bad ones) will be programmed to accept any valid answer.
Most often, of course, even if there are an infinite number of valid answers, one of them is going to be much more obvious than the others, and the vast majority of students will pick that one.
Why are you limiting a to the natural numbers?
Also, you have two sets, not a set of ordered pairs.
{a, b | a ∈ R, b = (-3/2)a}
I wasn’t making a complete statement of the entire solution, just an example. Your statement is certainly a. more comprehensive answer; I just didn’t have time and wasn’t in the mood to apply proper set notation.
Stranger
Maybe it would be more clear to the OP, as well as everyone, to consider the related question, “Describe all lines in the plane that intersect the point (3/2, 0).” Then it is geometrically clear what is going on. NB that for vertical lines you cannot write y as a function of x, so geometrically all the answers are parameterised by a projective line, something something
Would you mind telling us which “old high school algebra and trig” books, specifically, you found this problem, perhaps even with the name of the chapter/section in which it was found? The copyright date(s) of the textbooks could also be helpful.
I am not sure I understand what is meant by “solution set” and does the notation “3/2” mean a value of 1.5 or is it shorthand for, I dunno, perhaps those values of a and b where x would equal 3 and also those values were equal to 2?
Of course if the book is for Linear Algebra and the section is on matrix solutions, I am not going to be much help, I never understood matrices, neither in HS or College (and I had more math in college than some math majors).
I believe I can answer that one. Think of a and b as parameters. Given values for them, solve the equation ax+b=0 for x. The set of values of x making this an equation is called the “set of solutions”. Generically, the solution set will consist of a single value, but not always; for example when a=0.
The solution set would be the set of possible values of x that make the equation true. Basically, they’re asking the student to create a simple algebra problem whose solution is “x = 3/2”.
So, a=0 and b=0 would be a trivial answer, correct?
And a=2/3 and b=-1 would be a valid answer? Unless HS algebra has gotten extremely involved since the 1970s, I am not sure what more could be said.
I guess all values where b=-(3/2)a would also be correct, or would that be considered wrong for being too vague as it does not give specific values for a and b?
Not for the problem quoted in the OP, no. You will have additional solutions when a = b = 0, because any x will work.
Which is what makes it trivial, correct? The solution set is true, but so are all values outside the solution set.
I mean, when b=-3 and a=2, then you solve 2x-3=0 for x; you get x=3/2, or x\in\bigl\{\frac{3}{2}\bigr\} if you will. So far, so good. In the case a=b=0, the solutions are x\in\mathbb{R}, so no good. You want to determine all values of a and b which make the solutions x\in\bigl\{\frac{3}{2}\bigr\}, no bigger and no smaller.
Respectfully submitted for your perusal
From. . . “Fundamentals of Algebra and Trigonometry” by Earl Swokowski. Second edition. 1971. #20 on Page 51 
Ah, yes, and they’re getting at the fact that there are an infinite number of answers when they say “Are these the only possible values of a and b?”. And they’ve clearly dealt with the concept before, given that they have other, similarly-phrased questions.
I’m not questioning your answer, but it seems to me that is a bit much for HS algebra. At least, for the HS algebra they taught in the 1970s.
On review, I believe we are saying the same thing. The answer a=b=0 would have been marked wrong as being a trivial answer in any of my algebra, trig, or calculus classes I had in HS, but either a=2/3 and b=-1 or b=-3 and a=2 would both be acceptable. To me, the additional questions to the problem are vital to understanding.
Finally, looking at the information submitted below by the OP, it seems we are both correct. Incidentally, it is possible that is the text I used in HS.
It’s not wrong because it’s trivial; it’s wrong because it gives the wrong solution set.
The impression I got from the OP is pick a & b so that x = 3/2 is the only solution.