With absolute values, only the magnitude of the change matters. It doesn't matter if the change is positive or negative. So H must change more than TS.
So it seems to be (area of circle) * (circumference) * (1/2). It's less a question of why it doesn't work, than why your brother thought it would work. Maybe if you posted the derivation, we could point out the problem with it.
99 = 3 * 33, so 1 and 11 aren't the only values x or y can take.
There's 1 solution to your problem.
Don't be surprised that you're getting "ad hoc" methods, you've thrown together a few random arbitrary conditions, especially the "twin number" bit.
I think that's equivalent to \sum_{p=2}^{n} \frac{\left \lfloor n/p \right \rfloor}{p} , where the square brackets represent the floor function, and p runs through the primes less than or equal to n.
I don't know if that helps at all, and no doubt it can be simplified more so.
The generally accepted order of operations are BIDMAS (or BODMAS, same thing), which is:
Brackets, Indicies, Division and Multiplication, Addition and Subtraction.
So, brackets are computed first, then indices (powers), then division and multiplication have equal priority, and are done left...
There aren't any proofs if you keep on asking "why?", at some point or another you're going to have to accept things as true, without proof: these are axioms. Some common axioms for arithmetic are:
1) a + (b + c) = (a + b) + c
2) a + b = b + a
3) a + 0 = a
4) a + (-a) = 0
5) a*b = b*a
6)...
It's proven, after hundreds of pages, in this book:
http://en.wikipedia.org/wiki/Principia_mathematica
I definitely can't explain how it's done, it's all in symbolic logic, and I guess you start with a degree level mathematical education.
For the first equation, your first step is wrong. Where you've attempted to subtract x from both sides, you've actually subtracted x from the left side, but only \frac{x}{4} from the left side. Adding, for example, 2 to a fraction, is not the same as adding 2 to the numerator of the fraction. In...
Your lack of brackets or LaTeX is both annoying and confusing.
For your first question, the answer x < -3 only comes about if the equation is 2x + 5 < \frac{x-1}{4}, so I'm assuming that's what it is. From your working, it's a bit confusing as to what equation you're trying to solve...
It's certainly easy, from a practical point of view, to put in the values first, so you know if you're multiplying/dividing by a negative number, which would change the inequality sign. Mathematically, if you consider separate cases for each variable being positive or negative, it makes no...