Ok, so I understand that an equation is something like 15 = 15 , and that the only criteria as far as I can tell for it being an equation is that both sides are equal to each other.

I have a few questions, the first is a question about why two equations are related to each other? For instance, if I had an equation like 15 = 15 and I subtracted another equation 7 = 7 from it, then I would have 8 = 8, which is still an equation yes, and I understand that it is possible to get back to the original equation by adding 7 to both sides again, but is it directly equal to the original equation without algebraic manipulation?

The reason that I ask this first question is that I'm curious about how after algebraic manipulation a result from a newly gotten equation directly applies to the original. If I had x + 5 = 8, and I subtracted the equation 5 = 5 from it, I would get that x = 3. I would then find that I could plug that value into the original and find that 3 + 5 = 8, which is a true equation. But why does subtracting another equation allow me to find the unknown in the original?

And finally, what makes equations special and unique from fractions in that you can add, subtract, multiply, divide, etc, all you wish from both sides of an equation but you can't do that with fractions. I understand that when you do that with a fraction, the value changes, however, when you do that with an equation doesn't the equation change? why are equations special? why can you do freely whatever you want to an equation but not a fraction?

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    $\begingroup$ +1 I love this question. It's right at the boundary between what you understand and what you don't but want to understand. This boundary is exactly where all the action of learning takes place. Math.SE community, let's show nachos that they've come to the right place! $\endgroup$ – Ben Blum-Smith May 22 '14 at 15:59
  • $\begingroup$ If you look at my examples you will see that none of them are functions. x + 5 = 8 is an equation, but not a function, there is no independent -> dependent variable relationship going on there. There is only one possible for value for x that will map to the given value. So I don't know why people are bringing up functions. I wanted to know how two different equations are related, i.e 8 = 8 and 15 = 15 (see question for details), not how an equation is related to itself with functions. $\endgroup$ – nachos May 22 '14 at 16:45
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    $\begingroup$ The people who have answered in terms of functions see the transformation of one equation to another as applying a function to both sides of the equation. This is mathematically legit, but it's clear that it is not helping you, and I hope that others answer the question in ways that are more directly engaged with what's bothering you. $\endgroup$ – Ben Blum-Smith May 22 '14 at 17:23
  • $\begingroup$ Possibly related questions here and here. $\endgroup$ – Andrew D. Hwang May 23 '14 at 22:38

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).

  • $\begingroup$ thanks, after reading this and this post on Dr. math which information in your post lead me to: mathforum.org/library/drmath/view/53093.html, I think I finally get it. What causes a new equation to give an equivalent and thus usable answer to a previous equation which it was derived from comes from the fact that it is possible to undo the operations, or the "journey", that lead you from one equation to the other. They are not literally the same, but they are equivalent. $\endgroup$ – nachos May 25 '14 at 0:10
  • $\begingroup$ Yes, that's a good answer by Dr. Math, though I would caution that taking the square root isn't exactly the inverse of squaring (since $(-2)^2=4$ and $\sqrt{4}=2 \neq -2$). $\endgroup$ – Jedai May 25 '14 at 10:47

Ok, maybe the following formulation of my answer is more helpful:

An equation tells you that some quantity (variable, constant, expression built from variables and constants) is identical to another quantity.

You can now apply any operation to both sides (the same(!) operation to both sides) and this will preserve the equality. Because after all, if two objects are the same and you do the same thing to both objects, the results will be identical again.

For example if $x = 5y$ and you add $5$ to both sides, you arrive at $x + 5 = 5y + 5$.

This is not the same equation as the original one, but it is equivalent to the original equation, which (essentially) means the following: By applying some other operations on both sides of the new equation, you can get back to your original equation (in a sense, the operation that you applied is invertible).

In this case, you can subtract $5$ from both sides. This will lead to $(x+5) - 5 = 5y-5$, which leads to $x = 5y$ again.

Note that not all operations (like adding $5$ above) have the property that they yield equivalent equations (in the sense that you can "get back").

A striking example is the following: Consider the equation $x = y$. Now multiplying this equation by $0$ yields $0\cdot x = 0\cdot y$, which is fulfilled for ALL numbers $x,y$, whereas the original equation only holds for $x=y$.

This shows that you cannot "do freely whatever you want" to an equation and still get "the same" equation.

In summary, the following is true: Given an "original" equation, you can apply arbitrary operations to both sides and you will get new equations that are guaranteed to be fulfilled as long as the "original" equation you started with is fulfilled.

If you only used "equivalence operations" (i.e. operations in which you can "go back" by some other operation), the original equation will also be fulfilled as long as the "new" equation is fulfilled. (In this case, you can derive a "solution" to the "new" equation and this will also be a solution to the original equation).

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    $\begingroup$ I am shying away from a downvote, but I think that this answer is unhelpful to the OP. The OP has expressed discomfort regarding the relationship between $x+5=8$ and $x=3$. This indicates the need for engagement at a very fundamental level of understanding of what equations are (cf. the question title). Talk of functions and injectivity introduces concepts into the discussion that the OP is going to have to work hard to understand what they even have to do with his/her question. $\endgroup$ – Ben Blum-Smith May 22 '14 at 17:37

By definition, both sides of the equation are exactly the same. They may appear different, but they are the same. In the equation $x+3=8$ we mean that $x+3$ actually is $8$. We might be writing it in terms of this "unknown" $x$, but when it gets down to it, it's still $8$. So do whatever you want to both sides and you're doing it to the same thing $(8)$, so both sides must still be equal. $8^{12.6}+5$ is clearly the same as $8^{12.6}+5$, and this still holds true if I write one of those $8$'s as $x+3$, i.e. $(x+3)^{12.6}+5$, because we know that $x+3$ is actually (in this case) exactly the same as $8$.

Hope that makes sense :)


I think that the confusion comes from mixing concepts inappropriately. I just wanted to point out that you can't add equations. Consider $(15=15) + (-7=-7) = TRUE+TRUE$, but I don't know how to add 2 logical values together (but I do know how to AND them together). In this sense, a fraction and an equation are not comparable objects. When i add a value to a fraction, its value changes; likewise when I add something to the LHS (or RHS) of an equation, its value changes too (even though the relationship between the LHS and RHS is still one of equality).

The reason why $a=b$ implies that $a+c=b+c$ is because we assume it to be true, and no other reason, just as we assume that every number is equal to itself. It's fundamental to our very concept of numbers and equality. Another assumption that we make is that for all numbers (whether we mean Integers, Rationals (fractions), or Reals), the associative property is always true [$(a+b)+c = a+(b+c)$].

So if we assume that $$x+5=8$$ and that $$-5=-5$$ then we can also assume that $$(x+5)+(-5) = 8 + (-5)$$ and also assume that $$x+(5+(-5)) = 3$$ of course I did forget to mention that we also need to assume that $$a+(-a)=0$$ and that $$b+0=b$$ giving us $$x+0=3$$ $$x=3$$ And why can I now substitute $x=3$ back into the original equation ($x+5=8$)? It is because we have assumed that every number is equal to itself.


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