Question: I want to show that this *magic* equation is true for all kinds numbers (rational, irrational, imaginary, natural, etc..) which is true as far as I've tested.$$\displaystyle\large\frac{2x+10}{2}-x=5$$How would I exactly do that?

Answer: 
First, we have

$$\frac{2x + 10}{2} = \frac{2(x + 5)}{2} = x + 5$$

so the overall equation simplifies to

$$x + 5 - x = 5$$

which is obviously true for all $x$.
The proof is the same ("universal") in all the number systems that you mention simply because they all obey the *laws* that one employs to prove the equality (associative, commutative, distributive laws, etc). In addition to these [ring axioms,](http://en.wikipedia.org/wiki/Ring_%28mathematics%29) the proof also uses the fact that one can *uniquely* divide by $\,2,\,$ which is *not* true in every ring. For example, in modular arithmetic $$\rm mod\ 10\!:\ \ x\equiv1\ \Rightarrow\ \frac{2\,x+10}2-x\, \equiv\, \frac{2}2 - 1\,\equiv\, 0\,\not\equiv\, 5.$$

The problem is that $\rm\:mod\ 10\!:\ 2\, y \equiv 2\:$ has *two* solutions, $\rm\ y\equiv 1,6,\:$ hence the "fraction" $\rm\, y\equiv 2/2\,$ does not denote a unique value. Similar difficulties occur when attempting to generally apply the quadratic formula (which involves division by $2$). Analogous difficulties occur generally for fractions whose denominator is not coprime to the modulus. These matters are clarified when one studies abstract algebra (look up *zero-divisors* and *integral domains*).

In summary the "magic" is simply *abstraction*. By choosing ring axioms that abstract out the common *algebraic* structure of these familiar number systems, we are able to give a single proof that works universally for all number systems (rings). Thus we need prove only *once* familiar identities like the Binomial Theorem, Difference of Squares factorization, etc, since the proof can be specialized as need be for any particular ring (this specialization process will become even clearer *algebraically* when one learns about the *universal* properties of *polynomial* rings).