Two algebraic expressions are **equivalent**
if they always lead to the same result when you evaluate them,
no matter what values you substitute for the variables.
For example, if you substitute *x* := 3
in *x* + *x* + 4,
then you get 3 + 3 + 4, which works out to 10;
and if you substitute it in 2*x* + 4,
then you get 2(3) + 4, which also works out to 10.
There's nothing special about 3 here;
the same thing would happen no matter what value we used,
so *x* + *x* + 4 is equivalent to 2*x* + 4.
(That's really what I meant when I said that they mean the same thing.)

When I say that you get the same result,
this includes the possibility that the result is undefined.
For example, 1/*x* + 1/*x* is equivalent to 2/*x*;
even when you substitute *x* := 0,
they both come out the same (in this case, undefined).
In contrast, *x*^{2}/*x*
is *not* equivalent to *x*;
they *usually* come out the same,
but they are different when *x* := 0.
(Then *x*^{2}/*x* is undefined, but *x* is 0.)
To deal with this situation, there is a sort of trick you can play,
forcing the second expression to be undefined in certain cases.
Just add the words ‘for *x* ≠ 0’
at the end of the expression to make a new expression;
then the new expression is undefined unless *x* ≠ 0.
(You can put any other condition you like in place of *x* ≠ 0,
whatever is appropriate in a given situation.)
So *x*^{2}/*x* *is* equivalent
to *x* for *x* ≠ 0.

To symbolise equivalent expressions, people often simply use an equals sign.
For example, they might say
‘*x* + *x* + 4 = 2*x* + 4’.
The idea is that this is a statement that is always true,
no matter what *x* is.
However, it isn't really correct
to write ‘1/*x* + 1/*x* = 2/*x*’
to indicate an equivalence of expressions,
because this statement is not correct when *x* := 0.
So instead, I will use the symbol ‘≡’,
which you can read ‘is equivalent to’
(instead of ‘is equal to’ for ‘=’).
So I'll say, for example,

*x*+*x*+ 4 ≡ 2*x*+ 4,- 1/
*x*+ 1/*x*≡ 2/*x*, and *x*^{2}/*x*≡*x*for*x*≠ 0.

So far, I've only used an identity
to evaluate each expression for the same value of the variables,
that is to substitute constants (specific numbers) for variables
in the identity.
So for example,
if I substitute *a* := −3 and *b* := 4
into the commutative law for addition,
then I get −3 + 4 = 4 + (−3);
if you further remember that subtraction means adding the opposite,
this tells you how to calculate −3 + 4 as 4 − 3 = 1.
But in fact, identities are good for more than that,
and for Algebra we need to use them in more general ways.

The main idea is this:
Given any equivalence of algebraic expressions,
you can get another equivalence
by substituting (not necessarily a constant but)
*any* defined algebraic expression for one of the variables
(and you can do this multiple times to substitute for multiple variables).
For example, if in the commutative law of addition
you substitute *a* := −3 (as before)
but now also *b* := 2*x*,
then you get −3 + 2*x* ≡ 2*x* − 3.
This is also an example of the commutative law of addition,
just like −3 + 4 = 4 − 3 is,
but now it's an example that we'll need in Algebra.

First some terminology:
The **terms** of an algebraic expression
are the smaller expressions which are added to form the larger expression.
For example, the terms of 2*x* + 3 − *y*
are 2*x*, 3, and −*y*.
(Remember that subtracting *y* is the same as adding −*y*.)
The **factors** of a term
are the expressions which are multiplied to form that term.
For example, the factors of 2*x* are 2 and *x*.
The **coefficient** of a term
is the constant factor of that term
(or their product if there are several, or 1 if there is none),
together with the minus sign if there is one.
So the coefficients of the terms of 2*x* + 3 − *y*
are 2, 3, and −1.
Terms are **like** (or *alike*)
if they are the same *except* (possibly) for their coefficient.
For example, none of the terms above are like,
and the terms 2*x* and 2*x*^{2} are still not like,
but the terms 2*x* and −3*x* *are* like.

Remember that the distributive law (on the right)
says that (*a* + *b*)*c* ≡
*a**c* + *b**c*.
If you substitute the coefficients of two terms for *a* and *b*
and substitute their common factor for *c*,
then this distributive law (going backwards)
tells you how to combine those terms.
For example, if I substitute
*a* := 3, *b* := −5, and *c* := *x*,
then I get (3 − 5)*x*
≡ 3*x* − 5*x*;
since 3 − 5 = −2,
I can combine like terms
to get 3*x* − 5*x* ≡ −2*x*.

You can simplify any linear expression, no matter how complicated,
by repeatedly removing grouping symbols like this and combining like terms.
For example, consider the expression
5[3*x* + 4 + 2(*x* + 4)] −
7(*x* + 6*y* − 2*y*).
First, I combine any like terms inside the inmost grouping symbols,
then I get rid of those symbols by distributing the constant factor.
Next I combine any like terms inside the next grouping symbols,
then get rid of those symbols (the brackets) as well.
Finally, I combine like terms.
The final result is much simpler:

- 5[3
*x*+ 4 + 2(*x*+ 4)] − 7(*x*+ 6*y*− 2*y*) — original expression; - 5[3
*x*+ 4 + 2(*x*+ 4)] − 7(*x*+ 4*y*) — combine 6*y*and −2*y*into 4*y*; - 5[3
*x*+ 4 + 2*x*+ 8] − 7*x*− 28*y*— distribute 2 to*x*and 4, and −7 to*x*and 4*y*; - 5[5
*x*+ 12] − 7*x*− 28*y*— combine 3*x*and 2*x*into 5*x*, and 4 and 8 into 12; - 25
*x*+ 60 − 7*x*− 28*y*— distribute 5 to 5*x*and 12; - 18
*x*+ 60 − 28*y*— combine 25*x*and −7*x*into 8*x*; - 18
*x*− 28*y*+ 60 — rearrange terms (not really necessary, but sometimes nice).

A linear expression
can *always* be simplified to something like this:
a sum of a few terms,
one for each variable in the expression
(consisting of that variable multiplied by a constant coefficient,
or possibly just the variable or its opposite
if the coefficient is 1 or −1),
and one constant term,
except that even some of these terms might not show up
(if their coefficients happen to come out to 0).
Whenever you run across a linear expression,
you will almost certainly want to simplify it like this!

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