Well, to come back to the middle-school definition of an equation : an equation is an equality between two expressions which may contain unknowns (that is letters that represent quantities which we don't know yet). Solving an equation consist in finding all values by which we can replace the unknown and get a true equality. In this sense :
$$x + 3 = 8$$
is an equation (x is unknown) and its only solution is 5 because only $5+3$ equals $8$, not any other number plus 3.
Now that we have defined equations we may wonder how we can find their solutions in a mechanical way (and be sure that we found them all). To do this we know a certain number of way to find other equations that have the same solutions than the original (we say those equations are equivalent since they're true/false for the same values of the unknown), by choosing cautiously we may obtain an equation whose solutions are all obvious. Here by "subtracting 3" to both member of our first equation we get :
$$x + 3 = 8 \quad \Leftrightarrow \quad x + 3 - 3 = 8 - 3 \quad \Leftrightarrow \quad x = 5$$
And obviously, 5 is the only solution of this last equation... ($\Leftrightarrow$ means "is equivalent to" and often we make that implicit by putting equivalent equations on successive lines)
Now why do we get equivalent equations by subtracting 3 to each member ? Well that comes back to the fact that two equations are "equivalent" if they're true/false for the same values of the unknown : if you have two equal quantities and add/subtract the same amount to those two quantities, you'll still have equal quantities afterward, won't you ? And similarly if you start with different quantities, the results would still be different afterward.
That's why you can add or subtract the same number to both members of an equation when you want to solve it.
That's also why you can multiply or divide by the same non-zero number.
You can't multiply by zero because even if the initial equality was false it will becomes true : $10 \neq 5$ but $0 \times 10 = 0 \times 5$ since $0 = 0$.
There's a pretty simple criteria that dictate which operations you may use on both members simultaneously and obtains an equivalent equation : you can only use operations that you can reverse, that is there is an operation which allows you to come back to the initial value. "Adding 3" is reversed by "Subtracting 3", "Multiplying by 5" is reversed by "Dividing by 5" and so on.
$$7 + 3 = 10 \quad 10 - 3 = 7 \qquad -5 + 3 = -2 \quad -2 - 3 = -5 \qquad ...$$
$$3 \times 5 = 15 \quad 15 \div 5 = 3 \qquad -10 \times 5 = -50 \quad -50 \div 5 = -10 \qquad ...$$
Since there are a lot of operations with those characteristics, you can do a lot more than just add/subtract/multiply/divide by the same number but depending on your level you may not have seen any yet.
Note that there is a bit more to solving equations than I described here, in particular you may have cases where an equation is equivalent to several equations (and you'll need to be careful which logical operator you use in this case : and or or).
Rather, multiply the first equation by $\cos q$ and the second by $-\sin q$ and add these scaled equations to obtain $$x\cos(p+q)-y\sin(p+q)+z(\cos q -\sin q)=\cos q -\sin q+1.$$ Since from the last of the original triple we know that $x\cos(p+q)-y\sin(p+q)=2-z,$ we can substitute in the above sum and letting $K=\cos q -\sin q,$ we obtain $$2-z+Kz=K+1,$$ which you may now substitute in any pair of the original system to eliminate $z,$ so that you now have a pair in $x$ and $y$ alone. I believe you can find the way from there.
Best Answer
I think the difference you are picking up on has to do with the structure of the equation itself. In some equations, the unknown $x$ is acted on serially through a sequence of nested operations; solving the equation amounts to "unwinding" those operations one by one. For example, $\frac{(3x+5)^2-19}{2}=10$ can be thought of as "take $x$, triple it, add $5$, square the result, subtract $19$, and divide by $2$; the result is $10$. You can diagram this as a series of functions like so:
So to solve it, you reverse those steps: Start with $10$, multiply it by $2$, add $19$, take the square root, subtract $5$, divide by $3$.
But in other equations -- even fairly simple ones -- have a different kind of structure. For example, $x^2+5x=10$ looks like this:
Notice that this structure resists any attempt to solve it by "unwinding", precisely because of the fork in the diagram. The $x$ flows through more than one path, which makes it impossible to trace the result backwards to its source.
Some equations are presented in a form that at first appears to have multiple-paths (e.g. $3x + 5x = 16$) but we can rearrange them to a single-path structure (for example, via the distributive property / combining like terms). But higher-degree polynomial equations typically have terms that cannot be combined, and this is what makes them resistant to the kind of intuitive solution you are asking about.
Update: In the comments below, the OP asks:
Although completing the square is historically geometric in origin, it can be understood in a purely algebraic way as a method of restructuring a function so that it has a "serial" structure, enabling it to be solved via unwinding. Let's take the example of $x^2+5x=10$, already diagrammed above. In completing the square, you first add $(\frac{5}{2})^2=\frac{25}{4}$ to both sides of the equation, obtaining $x^2 + 5x + \frac{25}{4} = \frac{65}{4}$. Then you recognize the left-hand side as a perfect square trinomial, so the equation can be written $(x + \frac{5}{2})^2 = \frac{65}{4}$. This equation, if represented diagrammatically, would have a simple serial structure: Start with $x$, add $\frac{5}{2}$, square it, and end up with $\frac{65}{4}$. It can then be solved by unwinding: Start with $\frac{65}{4}$, take the square root(s) (keeping in mind that there are two square roots, one positive and one negative), and subtract $\frac{5}{2}$.
Of course, this is not the only method that can be used to tackle quadratics. Consider the slightly different example of $x^2 + 5x = 24$. If we rearrange this as $x^2 + 5x - 24 = 0$ and factor the LHS, we get $(x-3)(x+8)=0$. This equation can be diagrammed as follows:
At first glance this looks to be no better than the original equation; it has a fork in it, and seems resistant to unwinding. But! There is this property of real numbers, the "zero product property", which says that if two numbers multiply to be zero, then one of them must be zero. And that allows you to split the diagram into two separate diagrammatic cases:
And each of those can be tackled via a very simple unwinding method.
In short, most of the techniques that are taught (at least at the high school level) for solving polynomial equations can be understood as "methods for re-structuring equations so that unwinding techniques can be employed". (I'm not claiming that they are taught in those terms, or that they should be, but merely that they can be thought of that way.)
Unfortunately this only gets you so far. Once you get to 5th degree polynomials, it is a famous result that there may be solutions that cannot be expressed by a combination of "simple" operations (see here). That means, among other things, that there is no way to restructure a general 5th degree polynomial so as to enable a solution via unwinding techniques.