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?
[Math] Algebra: What allows us to do the same thing to both sides of an equation
algebra-precalculus
Related Solutions
My question is: Is there any disadvantages to thinking about Algebra like this? Is there anything later in my math education that will require me to know that I am subtracting or adding 2x to get rid of it on this side?
As a college algebra instructor, I'm involved with remediation efforts for hundreds of students each year who have graduated high school but can't get started with college math, mostly due to incorrect concepts picked up in their prior schooling. So I would say "yes". There are some shortcuts that teachers can take to get students to pass some specific tests or programs that they are involved in; but the incorrect concepts definitely make things more difficult for students, sometimes overwhelmingly so, later on. (A majority of students that land in college remediation programs never get college degrees.)
The first thing that I would point out is that the "apply inverse operations to both sides" idea is generalizable to any mathematical operation; this allows you to cancel additions, subtractions, multiplications, divisions, exponents, radicals... even exponential, logarithmic, and trigonometric functions. (With appropriate fine print: no division by zero, square roots to both sides creates two plus-or-minus solutions, trigonometric inverses creates infinite cyclic solutions, etc.)
In contrast, the "move over and change the sign" method is not generalizable, as it only works for addend terms. This sets students on a course that requires memorizing many apparently different rules, one for each operation, which is much harder. When solving $2x = 10$, how is the multiplier of 2 canceled out? Must we remember to move it and turn it into the reciprocal 1/2? Will the students mistakenly change the sign and multiply by -1/2? Or add or multiply by -2 (I see this a lot)? How do we remove the division in $\frac{x}{2} = 5$ (probably some other rule)? How will we remember the seemingly totally different rule to solve $x^2 = 25$?
By way of analogy, I have college students who never memorized the times tables; they did manage to get through high school by repeatedly adding on their fingers, and can get through perhaps the first part of an algebra course that way. But then we start factoring and reducing radicals: "What times what gives you 54?" I might ask; "I have no idea!" will be the answer (this happened this past week; and here's a student who has effectively no chance of passing the rest of the course).
In summary: There are shortcuts or "tricks" that can get a student through a particular exam or test, which prove to be detrimental later on, as the "trick" fails in a broader context (like in this case, with any operations other than addition or subtraction). This then sets a student on a road to memorizing hundreds of little abstract rules, instead of a few simple big ideas, and at some point that complicated ad-hoc structure comes crashing down. Be polite and don't fight with your teacher to change things; but make sure to pick up a broader perspective for yourself, and share it with other students if they're willing, because you will need it later on. Take the opportunity to think about how you could improve on teaching the material, and then you may be on the path to being a master teacher yourself someday, and helping lots of people who need it.
If you have 2x - 1 = -x + 1, you can simply add +x and +1 to both sides. This gives you 2x - 1 + x + 1 = -x + 1 + x + 1, which leads to 3x = 2 and therefore x = 2/3.
Best Answer
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
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