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 Friday 15th of December

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 Dependent Variable

 Number of inequalities to solve: 23456789
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# 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!).