# 1. Addition and Subtraction of Algebraic Expressions

Before we see how to **add and subtract integers**, we define **terms** and **factors**.

## Terms and Factors

A **term** in an algebraic expression is an expression
involving letters and/or numbers (called **factors**),
multiplied together.

### Example 1

The algebraic expression

5

x

is an example of **one** single **term**. It has **factors** 5 and *x*.

The 5 is called the **coefficient** of the
term and the *x* is a **variable**.

### Example 2

5*x* + 3*y *has **two** terms.

First term:5x, has factors `5` andx

Second term:3y, has factors `3` andy

The `5` and `3` are called the **coefficients** of the
terms.

### Example 3

The expression

`3x^2 - 7ab + 2esqrt(pi)`

has **three** terms.

First term:`3x^2` has factors `3` andx^{2}

Second term:`-7ab` has factors `-7`,aandb

Third Term:`2esqrt(pi)`; has factors `2`, `e`, and `sqrt(pi)`.

The `3`, `-7` and `2` are called **coefficients** of the
terms.

## Like Terms

"Like terms" are terms that contain the **same variables** raised to the **same
power**.

### Example 4

3*x*^{2} and 7*x*^{2} are **like terms.**

### Example 5

-8*x*^{2} and 5*y*^{2} are **not like terms**, because the variable is not the same.

## Adding and Subtracting Terms

**Important:** We can only add or subtract **like terms**.

**Why?** Think of it like this. On a table we have 4 pencils and 2 books. We cannot add the 4 pencils to the 2 books - they are not the same kind of object.

We go get another 3 pencils and 6 books. Altogether we now have 7 pencils and 8 books. We can't combine these quantities, since they are different types of objects.

Next, our sister comes in and grabs 5 pencils. We are left with 2 pencils and we still have the 8 books.

Similarly with algebra, we can only add (or subtract) similar "objects", or those with the same letter raised to the same power.

### Example 6

Simplify 13*x* + 7*y* − 2*x * + 6*a*

Answer

13*x* + 7*y* − 2*x * + 6*a*

The only **like terms** in this expression are `13x` and `-2x`. We cannot do anything with the `7y` or `6a`.

So we group together the terms we can subtract, and just leave the rest:

(13

x− 2x) + 6a+ 7y= 6

a+ 11x+ 7y

We usually present our variables in alphabetical order, but it is not essential.

### Example 7

Simplify −5[−2(*m* − 3*n*) + 4*n*]

Answer

Go back to Order of Operations if you are not sure what to do first with this question.

−5[−2(

m− 3n) + 4n]

The square brackets [ ] work just the same as round brackets ( ). We could have used curly brackets { } here as well.

The first thing we do is expand out the round brackets inside.

−2(

m− 3n) = −2m+ 6n

The negative times negative in the middle gives positive 6*n.*

Now add the 4*n* in the square brackets:

[−2

m+ 6n+ 4n] = [−2m+ 10n]

Remembering the −5 out front, our problem has become:

−5[−2

m+ 10n] = 10m− 50n

Taking each term one at a time, what we did was:

−5 × −2*m* = 10*m* (Two negative numbers multiplied together give a positive); and

−5 × 10*n* = −50*n* (Negative times positive gives negative)

Go back to the section on Integers if you are not sure about multiplying with negative numbers.

So here's the answer:

−5[−2(

m− 3n) + 4n] = 10m− 50n

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**Note:**

The fancy name for round brackets ( ) is "parentheses".

The fancy name for square brackets [ ] is "box brackets".

The fancy name for curly brackets { } is "braces".

### Example 8

Simplify −[7(*a* −
2*b*) − 4*b*]

Answer

−[7(*a* − 2*b*) − 4*b*]

= −[7*a* − 14*b* − 4*b*]

= −[7*a* − 18*b*]

= −7*a* + 18*b*

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