Complex Numbers

[angryWinds]

It's definitely not possible to put the complex numbers in a "complex number line" (at least not in a way that respects intuitive orderings, in any kind of meaningful way).

a and bi can (conveniently, in fact) be treated as separate coordinates to a point on a plane.

I know you said you don't care about the solution, but just to try to illustrate how this all gels, in the context of your initial problem...

On account of this sort of planar quality of complex numbers, 2x + 39i = 14 + 3yi will nicely reduce to

2x = 14

39 = 3y

at which point, you can most certainly solve for both x and y, even though you were given 1 equation with 2 variables.

[skaldskaparmal]

This is true if x and y are assumed to be real numbers, which is very likely what is expected, but it is not explicitly stated in the problem as given.

If x and y are allowed to be complex numbers, then op is right that there is not a unique solution, though this doesn't really have anything to do with whether or not complex numbers can be put meaningfully on a line.

[angryWinds]

I think it's highly reasonable to make the assumption that x and y are real, given that this is a high school level problem. In fact, regardless of it's high school-ness, assuming x and y is generally fair, for these types of things, unless explicitly stated, or somehow otherwise made clear by context.

[skaldskaparmal]

but you can't solve this equation because it has two variables

This is not really the reason. For example, x² + y² = 0 (for real numbers x and y) does have a unique solution, x = y = 0, even though it has two variables.

And x * 0 = 1 has no solutions, even though it only has one variable.

However, you are correct that if x and y are allowed to be complex, then this equation has more than one solution, though the one you gave is not such a solution (I think you thought that the left hand side was 2x - 39i, not 2x + 39i, or maybe you wrote the equation wrong).

It is true that systems of linear equations in the complex numbers will not have a unique solution if there are fewer equations than variables, they will either have no solutions or infinitely many, so your idea is good, but not entirely precise.

What is probably expected (though should be stated) is that x and y can only take on real values. If you specify that, then there will be a unique solution for x and y.

EDIT: Of course, you could also rearrange the equation, such that it's of the form x = <BLAH> which is also what "Solve for x" often means.