Algebra: What allows us to do the same thing to both sides of an equation? I understand that the expressions on both sides of an equal sign are the same entity, and I know that when you modify one side, the other must be changed because it is referring to the same thing. What I do not understand is why making a new equation (adding or taking away from an expression) allows one to know what an unknown represents. What about equations lets us do this?
 A: If you happen to be a more visual or manual thinker, try to visualize each equation as a set of masses on each side of a balance.  Numbers are marbles whose mass you know exactly, and variables are "blobs" whose mass at first is not known and which you are trying to find out.  The equals sign is the fulcrum of the balance.
For example, $5x + 3 = 18$ means that on one side of the balance you have five blobs and three marbles, and on the other side you have 18 marbles.  The balance is level.
Now, if you remove some marbles from one side, the balance will tilt to the other side, and if you add marbles, it will tilt to that side.  In order to keep things level, you must add or remove the same amount from both sides at the same time.
In the example, lets remove three marbles from each side.  That leaves 5 blobs on the left, and 15 marbles on the right, $5x = 15$.
Ok, it gets a little trickier for division.  Here, you want to think of taking the main balance and distributing the load over a group of smaller balances.  If you take the blobs off of the big balance and put one each on one side of five smaller balances, all will momentarily be tilted.  To bring them back to level, move marbles from the main balance to the smaller ones until everything comes back to level.  This will happen with three marbles on each small balance.  Now you have five small balances, each with one blob on one side, and three marbles on the other.  Thus $x = 3$.
I hope this verbal explanation is clear.  I can't do pictures from this computer, but if you like, I will try to add some over the weekend from another location.
A: It really comes down to the basic properties of equality: the reflexive property and substitution. 
Equality Property of Addition: Say $a = b$. Obviously $a+c = a+c$ (reflexive property). So $a+c = b+c$ (substituting $b$ for $a$ on the right side). 
Equality Property of Multiplication: Say $a = b$. Obviously $ac = ac$ (reflexive property). So $ac = bc$ (substituting $b$ for $a$ on the right side). 
And so forth.
A: symplectomorphic's answer is great but I'll just add in my point of view: since I'm a computer science student I'll make use of graphs (we all love graphs in computer science). So, suppose we have a graph in which nodes represent equations and edges represent transformations of those equations, as in functions used on both sides of an equation.
In terms of this graph, finding the solution means finding a way between the starting node (our original equation) and a node which yields an "obvious"$^1$ solution. In order for a solution to be valid, the way between it and the starting node has to be undirected. So every step, or transformation you make, you have to prove you're moving through an undirected edge. Let's do some examples$^2$. Suppose starting equation is $5x + 4 = 14$.
             (x + 0.8 = 2.8)-b-(x = 2)
            /a                 /a
(5x + 4 = 14)-----c----(5x = 10) 

Where $a(t) = \frac{t}5$, $b(t) = t - 0.8$ and $c(t) = t-4$, all are linear.
Another example
         (x^2 - x = 0) 
        /d
(x^2 = x)--e--(x = 1)

Where $d(t) = t - x$ (linear for $\forall x \in \mathbb{R})$ and $e(t) = \frac{t}x$ (linear for $\forall x \in \mathbb{R} - \{0\}$). As you can see, you still have to worry about assumptions.
Last example from symplectomorphic's answer:
(x = 1)-->(x^2 = 1)---(|x| = 1)

Ok this one's last for real (because graph ain't real if it doesn't have a cycle):
(x = 1)----(x + 1 = 2)
   |             |
(2x = 2)---(2x + 2 = 4)


$^1$ obvious here doesn't mean $x = a$, see example 2
$^2$this is undirected edge:
---

this is directed:
-->

A: What allows us to do the same thing to both sides of an equation?
It depends what thing is.  If thing is a function, then it is OK because that's the definition of function.  If thing is not a function, watch out.
Example:
$$
\frac{1}{2} = \frac{2}{4}
$$
...take the numerator of both sides...
A: Equations
Here's one way of looking at it: if $a=b$, then $f(a)=f(b)$, no matter the function $f(x)$.
From this point of view, solving an equation amounts to applying a sequence of functions in order to generate an equation whose solutions are easy to read off. What this sort of argument shows is that the new equations are logical consequences of the original equation.
So, for example, suppose you're given that $$x+1=2$$
and you want to subtract one from both sides. Then just apply the function $f(s)=s-1$, yielding $$x=f(x+1)=f(2)=1$$
Aren't we done? We've isolated $x$, so what more is there to do? The problem is that high school trains us to stop thinking once we get to this point. (Actually it trains us not to think about what we're doing at all and instead to rely on the process to do the thinking for us. But manipulating symbols, by itself, doesn't constitute a mathematical argument. Experts often omit the details, but that's because they know how to fill them in if they needed to; beginners should be taught how to fill in the logical details above and beyond the symbol manipulation.)
Indeed there are some subtleties here. First of all, just because this last equation is a consequence of the original equation doesn't mean the last implies the original. (That would amount to the very common mistake of thinking a conditional and its converse are logically equivalent.) In other words, for an arbitrary function $f(a)=f(b)$ need not imply that $a=b$: the operation you perform on both sides might not be reversible. (It was in the example I just gave because the function I applied was linear, and all (non-constant) linear functions have inverses that don't require domain restrictions, which makes the transformation "reversible." Unfortunately in school almost all the examples we start out with are linear, so we have our intuition about equation-solving trained on a very special set of examples, which don't illustrate what can happen in general.)
The failure of $f(a)=f(b)$ to imply $a=b$ explains why certain operations—for example, squaring both sides—might generate "extraneous solutions." I put quotation marks around that phrase, because it's something of a misnomer: they aren't actually solutions to the (original) equation, precisely because they're extraneous. So, for example, if you apply the function $f(s)=s^2$ to the equation $$x=1$$ you deduce that $x^2=1$. You could then apply the function $g(s)=\sqrt{s}$ to deduce that $|x|=1$. At this stage you could analyze the problem into cases (depending on whether $x$ is positive or negative) using the definition of absolute value and deduce that either $x=1$ or $x=-1$. But this doesn't mean that either answer is a solution to the original equation. (Obviously $x=-1$ doesn't satisfy the original equation!) That's because the second step, of squaring both sides, isn't reversible. The chain of implication doesn't flow all the way backwards.
Another subtlety is that applying a certain transformation to both sides may require you to make an assumption without even realizing it. In other words, some operations, such as dividing by $x$, tacitly carry certain restrictions. The function $f(s)=\frac{s}{x}$, for example, requires that $x\neq0$; otherwise the value of the function doesn't make sense. So if you have $$x^2=x$$ and you apply the function $f(s)=\frac{s}{x}$ to both sides, you're tacitly assuming that $x\neq0$. That's why in other cases you might lose solutions rather than generate extraneous ones.
Of course, not all equations have solutions. For example, applying $f(s)=s-x$ to the equation $$x=x+1$$
yields $0=1$. What this argument shows is that $$(\exists x)(x=x+1)\implies0=1$$
By contraposition we conclude that $$\lnot(\exists x)(x=x+1)$$
or in other words that there is no value of $x$ that satisfies the equation $x=x+1$, because assuming there is such a value leads us into a contradiction. (As this example illustrates, if we're being totally rigorous we should really pay attention to quantifiers. But that's more than you're asking.)
And some equations are, in fact, true for all values of the variables. Such equations are called identities. A silly example is
$$x=x$$
but a slightly more interesting example is
$$x^2-1=(x+1)(x-1)$$
If you try to solve identities like this one, you'll produce a tautology like $0=0$.
Inequalities
For what it's worth, you can also extend this idea to the logic of inequalities. If you apply a function $f(x)$ to the statement $a<b$, you'd typically like to conclude something like $f(a)<f(b)$ or $f(a)>f(b)$. In other words, you want to know whether the process preserves the direction of the inequality or reverses it.
But to draw such a conclusion, you generally need to know whether $f(x)$ is

*

*increasing, i.e., $a<b\iff f(a)<f(b)$ or

*decreasing, i.e., $a<b\iff f(a)>f(b)$
on the interval from $a$ to $b$. So, for example, $f(x)=x+2$ is always increasing, and $g(x)=-x$ is always decreasing, so applying $f$ to $a<b$ yields
$$a+2=f(a)<f(b)=b+2$$
but applying $g$ yields
$$-a=g(a)>g(b)=-b$$
This second fact is just what we mean when we say "multiplying or dividing both sides of an inequality by a negative number reverses the direction of the inequality." It's simply a consequence of the fact that the function $f(x)=-x$ is decreasing.
How about squaring both sides of an inequality? In that case we're dealing with the function $h(x)=x^2$. Well, $h(x)$ is decreasing on $(-\infty,0)$ and increasing on $(0,\infty)$, so you have to be careful about "squaring both sides" of an inequality. If $a<b$ and $b<0$, then $h$ is decreasing on $(a,b)$, so
$$a^2=h(a)>h(b)=b^2$$
but if $a>0$ then the inequality is reversed.
Summary

The moral of this story is:
when you do something to both sides of an equation or inequality, think carefully about what function you're applying. In particular, for equations, think about whether

*

*It's invertible on the same domain you start out with.


*Its application requires you to make any assumptions.
For inequalities, you also want to think about whether

*

*the function is increasing or decreasing on the interval defined by the inequality you're starting from.


A: To answer the last part of your question:
What about equations lets us do this?
The additive and multiplicative properties of equality.
Additive:
if a - b = c, then a - b + b = c + b, or a = c + b
Multiplicative:
if a/b = c and b not = 0, then b * a/b = b * c, or a = b * c
A: Think about it this way, it balances the equation. If you have some equation like 2x=10 and you want to find x, you can do anything you want to one side of the equation as long as you do it to the other one to balance it out. So I can divide out the 2 on the left but that means I have to divide the right side by 2 also. This gives me x=5. 
A: This is an extension to the other well said answers...
Well, in elementary, it is done to separate different types of things. And we separate them on either side of the $=$ operator of course.
So, take the following equation for instance:
$$x+3=9$$
Now, we know what is $3$, we know what is $9$, but we do not know what is $x$.
So, we want to find out what is $x$ equal to?
What we actually want to do is to give the above equation the shape of the following equation:
$$x=\text{a number which is called 6}$$
The idea of reversal of signs of numbers on changing their positions with respect to the holy $=$ sign might be a bit nonsensical for elementary students. So, we tell them this ingenious trick of achieving the same result.
So, instead of pushing the $3$ towards $9$ to make $x$ alone and find out what it compares to, we send a number counter to $3$ to remove it from there and add its effect on the other side.
Hope that helps...
A: The equals sign asserts a relationship between two expressions.  You said "entities", which I don't think is quite right; if you evaluate two equal expressions, the results will also be equal, and those might be called entities (or, an entity, for which we have two expressions), but I don't think it's helpful to think of the expressions as "entities".
If we modify an expression, we change what it evaluates to.  If we make the same modification to two expressions we can* preserve their relationship (e.g. equality) by changing what both of them evaluate to.  This does not depend on knowing what the expressions evaluate to; the change has the same effect when the expressions include unknowns.
For some expressions (I don't know how to name the category), a modification of the expression can be simplified to combine or remove knowns within the expression.  The way making a new equation (through a sequence of such modifications) can reveal what an unknown represents is by preserving the equality relationship while removing the knowns.  When all knowns are removed, one of the two equal expressions is the unknown, and the other expression evaluates to the same value as the unknown.

* Not all changes preserve all relationships; for example, when $a < b$, multiplying by $-1$ flips the inequality to $-a > -b$.  @symplectomorphic correctly points out the importance of sticking to operations that preserve the relationship and keeping track of conditions the operations impose or relax.
A: May I attempt to give an elementary answer, since you seem to prefer one?
You said: 

"...I know that when you modify one side, the other must be changed
  because it is referring to the same thing."

It is precisely the exploitation of this knowledge that allows us to know what an unknown represents.
When you look at an equation, say 
$$3 + 2x = 7$$
What do you think? You might think, "Ah, if only $x$ was alone on one side of the equation, I would be able to know what its value is! How can I make that happen?"
The answer is, of course, you make that happen by applying the property of equations that you already understand: do the same thing to the left that you would do to the right. 
"I wish that $3$ wasn't there. I want to remove it, but then I'll have to remove it from the right side as well."
$$2x = 7 - 3$$
"Oh, the $2$ is stuck in multiplication to $x$. I can remove it by dividing $x$ by $2$, but I would have to do the same thing to the right hand side as well." So, we get
$$x = \frac{7 - 3}{2} = 2 \,\,.$$
At each stage, you're just taking advantage of the concept of equality.
A: Instead of thinking of terms of numbers and operations and math garblygook, it may help to talk a real example.  Say you're a parent with a couple of twins. You want to treat them equally.
So if you give one a cupcake, you have to give the other a cupcake.  
If you smash one kid's cupcake, now you have to smash the other one.
If you cut one in half, same with the other, etc.
As far as how it relates to solving for an unknown, there's of course the assumption that the two sides are in fact equal to begin with... hence the '=' sign.
So in the case of our kids, if they run off and steal a bunch of jolly ranchers, but one of the kid hides some in a shoe box under his bed, you can deduce from the amount the other kid has how many he's hiding (like by taking away all of the ones you can see, and counting how many the other kid still has left).  In other words -- by doing "like" operations to both sides, you can eventually narrow down the problem into a plain "unknown = known".
Hopefully this is just another way to view math that helps with any confusion.   Symplectomorphic's answer is also very good and starts to address some more mind-bending concepts.
A: The statement $x=y$ means that $x$ and $y$ evaluate to the same thing, i.e. they have the same value.  One consequence of this is that the symbols $x$ and $y$ are interchangeable.  Thus in any mathematical expression, $x$ can be replaced with $y$ (or vice versa) without any change in meaning.  Thus the expressions $x-3$ and $y-3$ have the same meaning, as do the expressions $\dfrac{x}{40}$ and $\dfrac{y}{40}$, or even more complicated expressions like $\dfrac{x}{x+1}$ and $\dfrac{y}{y+1}$.  When two expressions have the same meaning and you evaluate them, they must evaluate to the same thing, i.e. they must be equal.  So therefore $$x-3=y-3$$ $$\dfrac{x}{40}=\dfrac{y}{40}$$ $$\dfrac{x}{x+1}=\dfrac{y}{y+1}$$
Of course, sometimes, expressions cannot be evaluated.  For example, to evaluate $\dfrac{x}{0}$ would be dividing by zero, so it cannot be done.  So in this case, (assuming again that $x=y$) although $\dfrac{x}{0}$ and $\dfrac{y}{0}$ have the same meaning, they cannot be equated, since they cannot be evaluated.  As symplectomorphic pointed out, the third example  $\dfrac{x}{x+1}$ cannot be evaluated if $x=-1$, so the equality does not hold in that case.
