Introduction to Algebra
January 21, 2005

What is Algebra?
Using Variables
Solving for Variables
Checking your Work
Some Example Problems

What is Algebra?

Algebra is where we substitute symbols for terms in an equation.  Instead of a four, we might have an "x".  Let's look at an equation for an example:

5 + x = 9

In the above example, the "x" is what we call a variable.  It's not really the letter "x" - if you add x to five, you don't get nine!  Instead, a variable is a placeholder symbol for a value that we don't know yet.  We should read the example equation above as:

"Five plus something equals nine"

When we say it that way, it's a little less confusing, isn't it?  In the case of the above example, that something is four.  5 + 4 = 9, right?  So, in this case, the variable has a value of four, or x = 4.

If one term is equal to another, we can substitute them for each other in an equation, right?  Look at the two equations below. 

4 + 10 = 14
4 + 5 + 5 = 14

These two equations come out to the same result: 14.  In the second equation, though, we subsituted 5 + 5 for 10. We can do this because they're equal: 5 + 5 = 10.  The answer will be the same if we substitute any terms that equal ten into that space in the equation. It's the same when we say that x = 4:  the two equations below are also the same.

5 + x = 9
5 + 4 = 9

So, a variable is a placeholder for a value that we don't know yet, because they are equal.  Variables are usually lower case x, y or z, but can be any letter.  Sometimes we even use special symbols for variables that have certain meanings, but we won't worry about those for now.

Using Variables

Now that we know what a variable is - why do we bother with them?  Let's see an example where variables would be used:

"I've already manufactured fifty seven pairs of pants.  The store has placed an order for one hundred and twelve, which I need to deliver friday.  How many more do I need to make by then?"

Let's take a look at the problem that we've outlined.  We know how many pants we have already, and we know how many we need.  What we still need to know is how many more pants we need to make.  Let's build the equation.

First, let's start with what we know.  We need one hundred and twelve pairs, so that will be our total.

_ + _ = 112

So far so good?  We also know that we have already made fifty seven pairs, so let's put those on the other side of the equation, as one of the terms we're adding to make up the total:

57 + _ = 112

Now we've run out of numbers that we know, but we still need to somehow represent the number of pants that we need to make, right?  We'll be adding that number to the number of pants we've already made to reach our goal of 112.  In this case, we can use a variable to represent that number!
First, a note:  it is important to define a variable when you create one.  Since variables just look like letters stuck into an equation, you need to let anyone who will see your equation know what the variable represents so that they don't get too confused.  We can do this with a let statement.  Now, we can finish our example:

Let x be the number of pants that still need to be manufactured.
57 + x = 112

There we go!  We've properly constructed our equation, with a defined variable.  We can read that equation as "Fifty seven pairs of pants plus some number of pants that still need to be made will equal a total of one hundred and twelve pairs of pants".  The only thing that we need to do now is to find out what value our variable has - how many pants still need to be manufactured?

Solving for Variables

Finding out what value a variable has is called solving for the variable.  In the above example, we'll be solving for x.

In order to solve for variables, let's go back to something we learned when we first introduced variables.  In an equation, we can substitute any terms that are equal to any other terms, right?  Remember that we not only can substitute x for 4 (and the other way around), but 5 + 5 for 10?
Take a look at the two equations below:

2 + 5 + 3	=	10
2 + 5 + 3	=	5 + 5

Wait a minute!  That doesn't look right!
The first time you see this, it may not - but it is right!  All that we've done is replace 10 with 5 + 5, which we decided we could do earlier on.  If you add up all of the terms on both the left side and right side of the equation, you still get 10 = 10.  In fact, as long as both sides of the equation equal one another, we can make all of the substitutions that we'd like!  When solving for a variable, we will need to do this quite a few times.

Let's look at another way of making substitutions, or balancing an equation.  Let's say that our equation is:

5 + 10 = 15

That seems simple enough, right?  Let's look at another equation that is very similar to the one given above:

10 = 15 - 5

That also seems simple enough, right?  Now pretend that instead of five, we'd used a variable (let's use x, which in this case x = 5):

x + 10	=	15
10	=	15 - x

Now that seems interesting, doesn't it!  The equation remains balanced if we're adding a value on one side or subtracting it from the other!  In fact, we can use this property in a very powerful way:  we can move terms from one side of an equation to the other by inverting the operation. 
What does that statement mean?  Well, if we're adding something on the left side of an equation, we can move it over to the right side and subtract it.  If we're dividing something on the right side of an equation, we can move it over to the left side and multiply it.

Multiplication and Division are inverses of each other
Addition and Subtraction are inverses of each other

Note:  if a term doesn't have an operation in front of it, then you assume it has a plus sign ("+"), because it is a positive number.  If it was a negative number, it would have a subtraction operator (or "negative sign") in front of it!

Let's see some examples (check the math for yourself!):
Note:  because "x" can be a variable, we use the "*" sign to mean multiplication.

4 + 12	=	16		4	=	12 / 3		15 + 3	=	9 * 2
12	=	16 - 4		4 * 3	=	12		(15 + 3) / 2	=	9

Okay - so now that we know how to move terms from one side of an equation to the other, how does that help us to solve for a variable?  Let's go back to our making pants example:

Let x be the number of pants that still need to be manufactured.
57 + x = 112

In order to solve for x, we are going to isolate it:  if we have only the variable (x) on one side of the equation, and terms that we can solve for on the other, we will know the value of the variable!  Let's take a look:

57 + x	=	112
x	=	112 - 57
x	=	55

By isolating x, we can do the math on the right side of the equation and see that x = 55!  We have solved for x, and found out that we still need to manufacture 55 pairs of pants.  If we can isolate the variable in any algebraic equation, we can solve for it.

Checking your Work

Before we can go through some examples of solving for variables, we have to learn how to check if we've made a mistake.  How do we do this?  By solving both the left side and the right side of the original equation, and making sure that they're equal.
Checking your work is very important with algebra:  not only will it show you if you've made a mistake, but can give you an idea of where you made the mistake instead of having to start over from the beginning!

To check your work, set up a little table:  The column on the left side should be headed by LS (for Left Side), the one on the right RS (for Right Side).  Then put the terms from the original equation into their proper columns, and simplify, substituting the values found for any variables as the first step.  Let's check the work on our pants example:

LS	|	RS
57 + x	|	112
57 + 55	|
112	|

We can see that we were correct in our example:  the numbers at the bottom of our left side and right side are the same. 

Some Example Problems

This all makes sense now, right?  Let's walk through some problems.

Problem #1
In this first problem, we're just solving for x. The only tricky bit is to make sure you divide the entire left side by the term on the right side.  See if you can follow along!

40 - 5	=	x * 7
(40 - 5) / 7	=	x
35 / 7	=	x
5	=	x

LS	|	RS
40 - 5	|	x * 7
35	|	5 * 7
	|	35

Problem #2
Solving for a variable when there are more than two terms isn't any more difficult:  just make sure to take care of them one at a time.

x * (-5) + 7	=	6 / 2 - 1
x * (-5)	=	6 / 2 - 1 - 7
x	=	(6 / 2 - 1 - 7) / (-5)
x	=	(3 - 1 - 7) / (-5)
x	=	(2 - 7) / (-5)
x	=	(-5) / (-5)
x	=	1

LS	|	RS
x * (-5) + 7	|	6 / 2 - 1
1 * (-5) + 7	|	3 - 1
(-5) + 7	|	2
2	|

Problem #3
Try this problem on your own!  Click here to see the solution, and don't forget to check your work!

 x / 3 + 2 = 81 / 9

Hopefully you now have an introductory understanding of algebra.  You can ask your teacher or parents for more problems.  Good luck!