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Balkan Terrapin or Western Caspian Terrapin (Mauremys rivulata), family Emydidae, found in the eastern Mediterranean regions of Europe and Asia

From the second equation we get that cheeseburger plus fries-squared is five. Subtracting cheeseburger, which is six, from both sides, we get that fries-squared is negative-one. Math fans will know that there are two solutions to this; either fries are the "imaginary unit" 𝒾 or they are its negative, -𝒾. We'll do the rest of the problem with 𝒾, keeping in mind that at the end we should also take the complex conjugates as solutions.

Finally, we have that cup to the power of fries, minus cup, equals three. Replacing fries with 𝒾, and moving a cup to the other side, we get that cup-to-the-𝒾 is equal to cup-plus-three.

Now, the weird part about this is the cup-to-the-i. The problem with this is that complex exponentiation is technically not a thing. That is to say, there is no one function which is mathematically equal to "input-to-the-power-of-𝒾". In fact, there are infinitely many such functions.

Fortunately, due to reasons that take about six pages to explain (trust me I've done it), there is one particular function that many people have agreed is "the most reasonable one". This is not a mathematical notion, but a human preference. Seeing as this question was presumably written by a human, I am comfortable with using this function.

So, what function is this? Well, given a complex number r∠θ written in polar form (if you don't know what that means don't worry), where -π < θ ≤ π, then (r∠θ)^𝒾 = e^(-θ)∠ln(r).

Applying this to our problem a value r∠θ will be a possible solution for cup if e^(-θ)∠ln(r) = r∠θ + 3. Splitting this into real and imaginary parts, we get two equations: e^(-θ) cos(ln(r)) = r cos(θ) + 3 and e^(-θ) sin(ln(r)) = r sin(θ). We can graph these equations on Desmos:

let the value of the glass = z, for z ∈ ℂ:

z^(±i) = 3+z

let z = r·e^(iθ) for r,θ ∈ ℝ, -π < θ ≤ π

∴ z = r·e^i(θ+2πn) for all n ∈ ℤ

∴ (r·e^i(θ+2πn))^(±i) = 3+r·e^i(θ+2πn)

distribute powers (apologies for the use of ∓):

r^(±i)·e^(∓(θ+2πn)) = 3+r·e^i(θ+2πn)

convert to the same base:

e^i(±ln(r))·e^(∓(θ+2πn)) = 3+r·e^i(θ+2πn)

split into real and imaginary components:

re: cos(±ln(r))·e^(∓(θ+2πn)) = 3+r·cos(θ)

im: sin(±ln(r))·e^(∓(θ+2πn)) = r·sin(θ)

The first problem is there's no super clear indication of what ring we're working on.

For example another valid solution I might accept would simply be "we are working in a ring mod in some characteristic prime p where p mod 4 is 1." There are other ways. In general any ring in which -1 is a square, we have good algebraic extensions. In fact I'd allow any ring where there's a good argument for why exponentiation works, and allow people to extend it where necessary.

For example, in mod 29, we can do the same thing and get fries equal to 12 or 17.

On the point of 'infinitely many functions that can represent input to the power of'... this depends. There are, for example, infinitely many possible ways of doing input multiplied by other input, but in the end we choose one out of wanting it to correspond nicely with some properties. Input to the power of has a lot of functions that satisfy the properties we usually want, but we ordinarily choose additional properties with good reasons for them so that there is only one. I mean yeah at some point it's mathematical convention but so are the properties we care about for, say, addition. And then yeah this chosen function, which is no more a choice of convention than choosing to work in ℂ is not injective on ℂ.

(To this sentiment, 'addition' of ordinals violates the usual sense in which we define addition, even if it satisfied the general intuition of addition, while there are alternative ways to add do not satisfy this intuitive notion, (see symmetric difference in the natural ring on the power set) .)

In general I feel that if we're being pedantic the problem is not sufficiently well defined to admit solution.

Now let's say we do actually restrict ourselves to ℂ, which is a reasonable thing to do for an arbitrary thing. Then most of this works.

It can also be found on Google Drive here and on Bluesky here.