If 5 g of cannabis indica and a pack of rolling papers together cost $110 and the weed costs $100 more than the papers, how much do the papers cost?

Correct! Believe it or not, most people would say $10...but if we let P be the price of the pipe, then P + 100 has to be the price of the weed, and the sum of the two is given as $110...so we may state: (100 + P) + P = 110 100 + 2P = 110 2P = 10 P = 5

Yes, that might be true. There is some rule of logic, from nonsense or contradiction anything may follow. "Let P be the price of X" is nonsense or an ideological use of mathematics. I mean you can believe that they are equal or unequal, but you can't demonstrate this mathematically using logical axioms and rules of inference since these are valid only for elements of the theory, prices of things in the mortal world are not elements of the theory. The ideological use is when you say the set 110 is the price $110.

Yeah now you're thinking of getting into quantum physics or something and that is waaaaaaaaaaay over nexopia's head lolz

I don't know...I thought I was pretty clear. If I prove that 1 + 1 = 2 in a mathematical theory, I am proving something about numbers which are not the prices of things in the mortal world, or try adding two drops of water, what do you get? So, trying to use that fact or a true statement of a mathematical theory to calculate the price of weed or rolling papers is nonsensical, I mean you can believe it if you want, but you can't demonstrate it.

In elementary algebra, a variable is a symbol we use to represent some unknown quantity. If we can construct an equation from known information involving our variable, then we may use algebraic operations/theorems to solve the equation for the unknown quantity. Certain equation types may require the use of numerical root finding methods (such as the Newton-Raphson method) to approximate a solution.

I think you mean set, natural numbers are sets, the integers are sets, the rationals are sets of integers, and the reals are sets of rationals. The variable ranges over the elements of a set, none of these elements are prices of things in the mortal world. Yes, we can prove statements involving variables and we can form sets of elements w/ a certain property (what you call solving equations). My point is all of this involves sets not prices and to say that sets are prices to convince someone is an ideological use of mathematics.

The concept of sets is not needed to solve the given problem. The fact that we obtain a linear equation with rational coefficients will naturally lead to a rational solution. Do we need to worry that the rationals are a subset of the reals which are a subset of the complex, etc.? No.

Yes, we don't need sets to solve or even pose the problem, since it is not a mathematical problem at all. A statement involving rationals is a statement involving sets, since rationals are sets. Yes, you are right, to solve an equation involving rationals we don't need to worry that rationals are sets, but that doesn't change the fact that they are sets. To solve your problem we don't need to worry that rationals are sets, because it is not a mathematical problem, it is ideological. Again, the only reason you believe this is a mathematical problem is because you believe that sets are prices, but you haven't demonstrated this mathematically.

No, what I "believe" is that a price is a numerical quantity, some quantity of monetary currency. This quantity (price) is typically a rational number (part of an arithmetic sequence where the common difference between each element is the quantum of the currency, in this case the cent), and the rational numbers are a set (not sets). So? Your objection is nonsensical, serving no use other than to be argumentative.

I once saw a pie at a bake sale by math students where the price was the pi number. They didn't know how to give me my change, lol.