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?

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    $\begingroup$ I take it that you already understand that if $a = b$, then $a - 3 = b - 3$ and $a/17 = b/17$ and so forth. You do not seem to have trouble believing such things; the question is, what use are those facts, and how do you figure out how to use them to get the answers you need? Is that what you want to know? $\endgroup$
    – David K
    May 23, 2014 at 2:13
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    $\begingroup$ Isn't it obvious? If quantity A is the same as quantity B, then if you do something to A, it is no longer the same as B, unless of course you do the same thing to B as well. For example if A = 5, and B = 5, then A + 1 = 6, and B+1 = 6. So if A = B, then A+1 = B+1 $\endgroup$
    – Kenshin
    May 23, 2014 at 2:58
  • $\begingroup$ Would "Because it works" be an unacceptable answer? $\endgroup$
    – corsiKa
    May 23, 2014 at 16:20
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    $\begingroup$ possible duplicate of Why do we have to do the same things to both sides of an equation? $\endgroup$ May 23, 2014 at 17:17
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    $\begingroup$ Because the equals sign means that each side of the equation represents the same number. If you do the same thing to the same number, it will come out the same. $\endgroup$
    – Alexander Gruber
    May 23, 2014 at 20:06

12 Answers 12



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 $0=0$.


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


but if $a>0$ then the inequality is reversed.


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.
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    $\begingroup$ This is great and all but do you really think it's helpful to someone who is asking an elementary question? $\endgroup$ May 22, 2014 at 22:48
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    $\begingroup$ @KingSquirrel: yes. I would have much preferred that someone taught me this in high school. All the core of my answer requires by way of prerequisites is an understanding of functions, which you have no evidence the OP lacks. $\endgroup$ May 22, 2014 at 22:50
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    $\begingroup$ No, what you're claiming is that an elementary question should be answered in intuitive terms. That depends entirely on context, and in the absence of context, I supplied a mathematical explanation. Your answer isn't mathematical: it's metaphorical. That's fine, but you can't know the OP didn't want something more rigorous. $\endgroup$ May 22, 2014 at 22:53
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    $\begingroup$ @Cole: any function that isn't injective will do. I cited the squaring function. I can't cite them all. ;) $\endgroup$ May 22, 2014 at 23:54
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    $\begingroup$ Wow, this is precisely how this subject should be taught! They instead just teach us that "you can do something to both sides and it works out". And then later you get introduced to "special cases" like square rooting where you have to be careful. I suppose it ends up making sense intuitively but much better to have a rigorous explanation like this. $\endgroup$
    – Claudiu
    May 25, 2014 at 13:50

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.


$$ \frac{1}{2} = \frac{2}{4} $$

...take the numerator of both sides...

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    $\begingroup$ Haha very nice. To make the distinction even clearer you could say that you can do anything to the value of both sides. "Taking the numerator" is doing something to the expressions on both sides, which has little to do with the value (usually). $\endgroup$
    – user21820
    Oct 31, 2015 at 10:08
  • $\begingroup$ I've always been cautious before applying a function on both sides, even though like u said it should be OK to do it. It is in the definition of a function after all (ur answer made me remember that). My issue has probably always been overlooking the details of a function's definition (like the sets it is defined over) before applying it. $\endgroup$
    – KraZZ
    Jul 26, 2021 at 11:09

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.

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    $\begingroup$ how are you going to use this picture to solve $x^2=x$ or $\sqrt{x}=5$? it's a nice metaphor for linear equations; it's not so helpful for anything else. $\endgroup$ May 24, 2014 at 16:48
  • $\begingroup$ @symplectomorphic: I agree 100%. But if it gets a learner started and helps with the basics before moving on to more advanced topics, what's the harm? $\endgroup$
    – cobaltduck
    May 24, 2014 at 17:06
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    $\begingroup$ the harm is that you haven't been explicit that your picture has serious limitations. absent such caveats, beginners could think your picture is supposed to apply in general. (cf. my remark in my answer on having our intuitions trained on very bad examples that don't illustrate what happens in general.) $\endgroup$ May 24, 2014 at 17:08

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.

  • $\begingroup$ It allow us to go from $x=1$ to $x*x=x$, where second one has extra solution of $x=0$. $\endgroup$ May 30, 2023 at 9:29

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) 
(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:

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    $\begingroup$ this is a good way of visualizing the transformations: it comes close to commutative diagramming what's going on, and I teach this way, too. I especially like that you can see visually that there are many "paths" to the same solution, even for boring linear equations. but you'd have to be careful with $e(t)=\frac{t}{x}$. the edge there isn't really undirected, precisely because of the tacit assumption that $x\neq 0$. $x=1$ implies $x^2=x$, but not the other way around. $\endgroup$ May 24, 2014 at 14:19

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

  • $\begingroup$ Wiki en.wikipedia.org/wiki/Equality_(mathematics) properties are called Substitution and Reflexive. However, Substitution properly applied to multiplication allow us to go from $x=1$ to $x*x=x$, where second one has extra solution of $x=0$. $\endgroup$ May 30, 2023 at 9:22

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.

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    $\begingroup$ Is there something in particular that allows me to do anything to both sides as long as it is equal? Am I just over thinking this? $\endgroup$
    – aaax2178
    May 22, 2014 at 23:08
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    $\begingroup$ No. It's just that when you have something equalling something else, you can do whatever you want to one side but to even everything out, you must do it to the other side. $\endgroup$ May 22, 2014 at 23:14
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    $\begingroup$ While I understand the effort of trying to be as simple as possible, the old Einstein quote "Everything should be made as simple as possible, but no simpler" applies here too I think. For example this does not explain why $x = x^2$ misses one solution if we just divide by x to get $x = 1$.. $\endgroup$
    – Voo
    May 22, 2014 at 23:44
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    $\begingroup$ @Voo It does not miss a solution. We couldn't have divided both sides by $0$ in the first place. So by dividing both sides by $x$, we must assume $x\neq 0$. Checking the case $x\neq 0$ and dividing both sides by $x$ brings us the solution $x=1$ and checking the case $x=0$ proves that it is a valid solution as well. $\endgroup$
    – user26486
    May 23, 2014 at 8:51
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    $\begingroup$ @mathh Yes I know, the point is: The underlying idea of why we have to check for 0 particularly isn't explained. And while "division excludes 0" is a simple obvious example (that's why I picked it after all), there are other situations where it's not as obvious. And that we can easily generate superfluous results as well is something that many beginners tend to overlook. Hence the general idea of what goes on is important I find. $\endgroup$
    – Voo
    May 23, 2014 at 19:42

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:


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...


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.


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.


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.

  • $\begingroup$ your third example also requires that neither $x$ nor $y$ are equal to $-1$. $\endgroup$ May 24, 2014 at 15:22
  • $\begingroup$ I edited the answer $\endgroup$ May 24, 2014 at 15:30

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.

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    $\begingroup$ This is a mathematics forum. Please don't refer to "math garblygook." (The word you're looking for is "gobbledygook.") $\endgroup$ May 23, 2014 at 11:47
  • $\begingroup$ Thank you for explaining why you downvoted. I find it interesting that somehow my net-reputation with this answer is positive. I'm not a SE guru by any means, but wouldn't that mean it was upvoted? I can only assume that your comment has caused people to have an emotional response to downvote based on a word rather than the point of the answer -- which was precisely to provide an answer that didn't involve mathematical syntax. Thanks. $\endgroup$
    – emragins
    May 24, 2014 at 16:01
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    $\begingroup$ I didn't downvote your answer, and I can't control how other people vote. nevertheless you offered only a metaphorical explanation (with so much detail -- e.g., stealing jolly ranchers -- that it isn't clear what's essential to the comparison), and you disparaged a mathematical explanation by calling it "garblygook." this forum is for supporting mathematical thinking, not criticizing it as nonsense. just because the question was an elementary one doesn't mean it doesn't deserve a careful, precise explanation. $\endgroup$ May 24, 2014 at 16:06
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    $\begingroup$ (continued) you're encouraging the attitude that there is something inherently misleading or intimidating about mathematical syntax. that is not an attitude worth promoting. surely it can obscure the heart of the matter: it would have if I'd written everything in the language of first-order logic and not explained anything in everyday English. but syntax can also be useful, a tool for capturing ideas precisely and concisely. $\endgroup$ May 24, 2014 at 16:16
  • $\begingroup$ I agree that there is the other side of the coin -- unfortunately I was (and frankly still am) pressed for time. Syntax is most definitely useful, but it can also be intimidating and often obscure the heart of the matter. (I'm reminded of many intuitive concepts that are made daunting due to a formulaic approach, such as the distance formula or even computing slope.) I suppose I see syntax as a tool, but I prefer when the understanding comes first followed by an explanation of how syntax can represent that understanding (which my answer is admittedly void of.) $\endgroup$
    – emragins
    May 24, 2014 at 16:23

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