Equivalent expressions

We often want to replace a complicated expression with a simpler one that means the same thing. For example, the expression x + 4 + 2 obviously means the same thing as x + 6, since 4 + 2 = 6. More interestingly, the expression x + x + 4 means the same thing as 2x + 4, because 2x is x + x when you think of multiplication as repeated addition. (Which of these is simpler depends on your point of view, but usually 2x + 4 is more convenient in Algebra.)

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 2x + 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 2x + 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, x²/x is not equivalent to x; they usually come out the same, but they are different when x := 0. (Then x²/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 x²/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 = 2x + 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 ≡ 2x + 4,
1/x + 1/x ≡ 2/x, and
x²/x ≡ x for x ≠ 0.

The textbook, however, just uses ‘=’ for everything, so you can too, if you want.

Equivalences and identities

All of the identities that I mentioned last week (and which are listed in the textbook on page 78) are really equivalences of algebraic expressions. For example, last week I mentioned the commutative law of addition: a + b = b + a for any real numbers a and b. But to say that these are equal for any real numbers is simply to say that the expressions are equivalent: a + b ≡ b + a. For a more complicated example, consider the multiplicative inverse law (on the right): a(1/a) = 1 for any nonzero real number a. In an equivalence, we need to note that a cannot be zero, so it becomes a(1/a) ≡ 1 for a ≠ 0. In this way, every identity gives an equivalence, and similarly every equivalence gives an identity.

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 := 2x, then you get −3 + 2x ≡ 2x − 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.

Combining like terms

Probably the most important application of an identity in Algebra is the application of the distributive law to combine like terms.

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 2x + 3 − y are 2x, 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 2x 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 2x + 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 2x and 2x² are still not like, but the terms 2x and −3x are like.

Remember that the distributive law (on the right) says that (a + b)c ≡ ac + bc. 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 ≡ 3x − 5x; since 3 − 5 = −2, I can combine like terms to get 3x − 5x ≡ −2x.

Simplifying linear expressions

Recall that linear expressions use mostly addition (and subtraction) and include multiplication only by constants. The key to simplifying linear expressions is thus the distributive law on the other side: a(b + c) ≡ ab + ac. Now a will be a constant, while b and c will be arbitary linear expressions. For example, 2(x + y) ≡ 2x + 2y; perhaps more interestingly, −5(3x − 4) ≡ −5(3x) − 5(−4). This latter example can be simplified further by multiplying the constants involved; so −5(3x − 4) ≡ −15x + 20 in the end. (Notice how (−) · (−) becomes +.)

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

So in the end, 5[3x + 4 + 2(x + 4)] − 7(x + 6y − 2y) ≡ 18x − 28y + 60.

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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