Collecting Like Terms
In an algebraic expression, there are terms with identical
symbolic or variable parts. We call them like terms.
For example,
‘4x’ and ‘2x‘ are like terms because both
of them contain the symbolic part ‘x’
‘3xyz^{ 2}’ and ‘11x yz^{ 2}’
are like terms because they both contain the symbolic part
‘xyz^{ 2}’
‘2x^{ 5} ’and ‘2x’ are not
like terms because even though the symbol present in both is
‘x’, the symbolic part ‘x^{ 5} ’ is
not identical to the symbolic part ‘x’.
We can simplify algebraic expressions containing like terms by
combining each group of like terms into a single term.
In fact, it is quite easy to see why this is possible (and
valid!). For instance, consider the expression
2x + 5x
which is the sum of two like terms, representing the
accumulation of two x’s and another five x’s. It is no
news that the result of this sum is 7x.
2x + 5x = (2 + 5)x = 7x
We can combine (or collect) like terms with each group of like
terms that appear in an expression. The result will be an
equivalente expression to the given one, which looks simpler (in
the sense that it can be written in fewer terms).
This process of combining (or collecting )
like terms can be performed for each group of like terms that
appear in an expression. The net effect will be that the original
expression can now be written with fewer terms, yet which are
entirely equivalent to the terms in the original expression.
Sometimes a bit more care must be taken to recognize like
terms.
Example:
Simplify 5abc – 3bca + 9cba – 4cab
solution:
This expression has four terms altogether. At first glance,
all four may seem to have different literal parts, since if we
were reading the literal parts as English words, abc, bca, cba,
and cab would certainly appear to be different.
However, keep in mind that these literal parts of the terms
are not characters in a word, but each character or symbol
represents a number. A literal part such as ‘abc’ then
represents the result of multiplying together the three numbers
represented by ‘a’, ‘b’, and ‘c’.
When we multiply three numbers together, it doesn’t matter
in what order we do the multiplication:
3 Ã— 4 Ã— 5 = 4 Ã— 3 Ã— 5 = 5 Ã— 4 Ã— 3, … etc.
Similarly, then
abc, bac, cba, and cab
really represent the same result or quantity because they are
products of the same three numbers, but with the multiplications
being done in different orders.
An easy way to be able to compare the literal parts of the
terms in the example expression here is to write the symbols in
each product in alphabetic order. Thus
5abc – 3bca + 9cba – 4cab = 5abc – 3abc + 9abc
– 4abc
Now we see that all four terms here have exactly the same
literal parts, and so all four are like terms. Thus,
5abc – 3bca + 9cba – 4cab = 5abc – 3abc + 9abc
– 4abc
= (5 – 3 + 9 – 4)abc
= 7abc
So, the original four term expression can be simplified to
just ‘7abc’. No matter what values are substituted for
a, b, and c, the simplified expression will give the same result
as the original, much lengthier expression (which is what we mean
by saying that the two expressions are equivalent). Obviously
then, it is preferable to retain the simplified expression, 7abc,
in place of the original, much lengthier expression.
Some care must be taken to ensure that terms being combined in
this was really are like terms.
Example:
Simplify: 5x^{ 2} y – 3xy^{ 2} + 7yx^{
2}
solution:
All three terms in this expression involve x, and y, and a
square. Following the strategy used in the previous example, we
rewrite the expression so that the symbols in each term appear in
alphabetic order:
5x^{ 2} y – 3xy^{ 2} + 7yx^{ 2} =
5x^{ 2} y – 3xy^{ 2} + 7x^{ 2} y
Now you can see that there are two like terms here – the
first and the third. The second term, despite involving the same
symbols, one squared and one not, is not actually
“like” the other two because the symbol which is
squared is not the same one:
x^{ 2} y is not the same as xy^{ 2}
Thus, all we can do here is to combine the first and the third
terms, which are like terms, to get
5x^{ 2} y – 3xy^{ 2} + 7yx^{ 2}
= 5x^{ 2} y – 3xy^{ 2} + 7x^{ 2}
y
= (5 + 7)x^{ 2} y – 3xy^{ 2}
= 12x^{ 2} y – 3xy^{ 2}
as a final answer (for now!).
