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Variables, Explained

Date: 05/05/2003 at 19:07:33
From: Vinnie
Subject: Algebra

I have problems trying to understand how to answer and work out these 
types of math problems:

  13.7b - 6.5 = -2.3b + 8.3

Date: 05/05/2003 at 19:25:08
From: Doctor Ian
Subject: Re: Algebra

Hi Vinnie,

Is it the variables that are troubling you, or the use of decimals? 
For example, would you be able to solve this? 

  14b - 6 = 2b + 1

Give that a try, and let me know what you come up with.  (Show me 
your work, too, since that will help me figure out how we can get 
to the next step together.)  

- Doctor Ian, The Math Forum 

Date: 05/05/2003 at 19:38:29
From: Vinnie
Subject: Algebra

First I'd like to thank you for taking time on reading and replying to  
my message.

Lets see...

14b - 6 = 2b + 1

     8b = 2b + 1
     8b = 3b

I am not sure I understand very well how to answer these types of 
problems. What would be the first step?

Date: 05/05/2003 at 22:46:18
From: Doctor Ian
Subject: Re: Algebra

Hi Vinnie,

Thanks for getting back to me, and showing your work. That really 

It looks like the first thing you're having trouble with is what it
means when we write something like 

  14b - 6

The b is a variable. That is, it stands for some number whose value
we're trying to find. When we write a number and a variable, or a
number of variables, together without any operator, multiplication is
implied. So '14b' means

  14 times (whatever b is)

If the value of b turns out to be 3, the value of '14b' is 

  14*3 = 42

If the value of b turns out to be 5, the value of '14b' is 

  14*5 = 70

Does that make sense?  So when we have something like 

  14b - 6

we can't really do the subtraction, because we don't know what the
value of b is.  

So let's look at a very simple equation, like

  3b + 2 = 11

This says:

  There is some number whose value we don't know yet.  
  Call that value 'b'. 
  If we multiply b by 3, and add 2 to the product, we get 11.

Usually the goal in a case like this is to find the value of b that
makes the equation true. We could try to find it by guessing:

  Does b = 0?

  3*0 + 2 = 11

    0 + 2 = 11

        2 = 11

This is a false statement, so the value of b is _not_ zero.  

  Does b = 1?

  3*1 + 2 = 11

    3 + 2 = 11

        5 = 11
This is also a false statement. Is there something we can do that's 
quicker than trying possible values? (What if the correct value turns 
out to be 1058? That will take us a long time to find!)

We can reason this way.  Looking at 

  3b + 2 = 11

we can think of it as 

  something + 2 = 11

which means that 

      something = 11 - 2

                = 9

Does that make sense?  Then since our 'something' is just 3b, we know 

             3b = 9

Now, we can think of _this_ as 

    3*something = 9

which means that 

      something = 9/3

                = 3

But our 'something' is just b. So we know that b=3. Let's check that 
by substituting 3 for b in the original equation:

   3b + 2 = 11

  3*3 + 2 = 11

       11 = 11

And this is true. So we've found the value of b that we were looking 

Were you able to follow all this? If so, then try to find the values 
that make the following equations true:

   5x - 4 = 1

   4x + 7 = 23

   3x + 5 = 2

Let me know what you get, and then we can go to the next level of
complexity, okay?

- Doctor Ian, The Math Forum 

Date: 05/06/2003 at 09:15:01
From: Vinnie
Subject: Algebra

I think I'm following. You are trying to find the missing number 
that makes the things in the equation equal, right?

  5x - 4 = 1

Here x is nothing because 5-4 is already 1. So the x confused me here, 
unless x = 1. Then I would understand because 5(1)-4 = 1, so x equals 
1, right? 

    4x + 7 = 23
  4(4) + 7 = 23  ----> x = 4 because 4*4 + 7 = 23 
Did I get it?

  3x + 5 = 2

In this one I don't understand how you can get it equal to 2.
But so far so good. I understand what you mean.

Date: 05/06/2003 at 10:44:33
From: Doctor Ian
Subject: Re: Algebra

Hi Vinnie,

You're right, we _are_ looking for the number that makes each equation 

It's a little like what we could do with sentences. Suppose we make a 
sentence with a blank in it:

  _______ is President of the United States.

There are lots of things we could put in the blank that would make 
the sentence false:

  Bob Dole is President of the United States.
  Homer Simpson is President of the United States.
  Jennifer Lopez is President of the United States.

But there is at least one thing we could put in to make it true:

  George Bush is President of the United States.
A variable is like that. You can think of it as a 'hole' or a 'blank'
in a sentence involving numbers. So when we write 

  5x - 4 = 1

we mean

  5*___ - 4 = 1

and we want to find what goes in the blank to make the sentence true. 

Why use letters instead of blanks?  Mostly because there are times
when we need to fill in more than one blank, e.g., 

  ___ is President of the United States, and ___ is Vice-President.

and we want to be able to tell the different blanks apart, so we give
them names. And being lazy, we make the names as short as possible -
usually a single letter:

  X is President of the United States, and Y is Vice-President.

  Y is older than X.

Note that this only works if we have

  X = George Bush
  Y = Dick Cheney

It doesn't work if we assign them the other way.  

If it helps, you can rewrite single letters as words, if that will
make it easier to remember what role they're supposed to be playing. 
That is, when someone writes

  5x - 4 = 1

you could immediately rewrite that as

  5*something - 4 = 1

Eventually, you'll get used to thinking of letters as representing
unknown quantities, but there's no law of nature that says an unknown
quantity has to be represented by a single letter, or that if you 
use a single letter it has to be x. 

>  5x - 4 = 1
>Here x is nothing because 5 - 4 is already 1. So the x confused me 
>here, unless x = 1.  Then I would understand because 5(1)-4 = 1, so x 
>equals 1, right? 

That's exactly right. The assignment x = 1 makes the equation true, 
and any other assignment (e.g., x = 3) makes the equation false.  

>    4x + 7 = 23
>  4(4) + 7 =23  ----> x=4 because 4*4+7=23 
>Did I get it?


>  3x + 5 = 2
>In this one I dont understand how you can get it equal to 2.

This one was tricky.  In this case, the value has to be a negative
number. Here's one way you might think of it:

         3x + 5 = 2         

  something + 5 = 2

         -3 + 5 = 2         so something = -3

             3x = -3

  3 * something = -3        so something = -1

              x = -1

If we put the value back in the equation, we get

  3(-1) + 5 = 2

     -3 + 5 = 2

          2 = 2

which checks.  

I put the negative case in there to make a point. In algebra, you
start out with equations where we can often find the right value by
making a few guesses. But quickly we get to equations where there
might be zillions of possible guesses - for example, where the value
we're looking for could be large or small, positive or negative, an
integer or a fraction or even an irrational number.  

For those equations, it's helpful to have some rules that we can
follow to help us proceed from something that looks like 

  16b - 6 = 2b + 1

to something that looks like 

        b = 1/2

And for the most part, there are really only two rules. The first is
that we can never, ever, for any reason, divide by zero. The second
is that whatever we do to one side of an equation, we can do to the
other side without changing the truth of the equation. 

For example, if we have an equation like

  16b - 6 = 2b + 1

this tells us that, for some value of b, the quantity on the left is
equal to the quantity on the right. Does that make sense?  Well, if
we have two things that are equal, and we add 4 to each of them,
they'll still be equal, right?  So let's do that:

  16b - 6 + 4 = 2b + 1 + 4

If we multiply both of them by 11, they'll still be equal, right?

  11(16b - 6 + 4) = 11(2b + 1 + 4)

Now, it turns out that most things like this that we might try aren't
all that helpful. They just make things more complicated. But in any
given situation, there will usually be a few adjustments that will
help us turn what we have into something simpler. For example, if I
start with 

  16b - 6 = 2b + 1

and add 6 to both sides, I end up with something simpler:

  16b - 6 + 6 = 2b + 1 + 6

      16b + 0 = 2b + 7
          16b = 2b + 7

That's simpler than what we started with. We can do something similar
by subtracting 2b from each side. (We don't know what b is, but it's
some number, and so 2b is also a number, and we can subtract the same
number from each side.)

    16b - 2b = 2b - 2b + 7

Now, on the right side, whatever b is, 2b minus 2b is zero. So the
right side simplifies to 

    16b - 2b = 0 + 7

    16b - 2b = 7

Now, what about the left side? To simplify the left side, you need to
use the distributive property of multiplication over addition, which
sounds kind of complicated, but it's actually much simpler than its
name would imply; and in algebra, it's one of the best friends you can
have. If you're not familiar with the distributive property, take a
moment to read this:

   Distributive Property, Illustrated 

Let's try to use the distributive property on the left side of our

  16b - 2b = 7

First, we find a factor that they have in commmon.  How about b?

  b(16 - 2) = 7

Make sure you understand what I did here, because it's one of the keys
to solving just about any algebra problem.  Anyway, now we can do the

      b(14) = 7

        14b = 7

Now, eventually you'll learn to look at things like 

  16b - 2b

and immediately simplify that to 


but it's good to know _why_ you can do that, i.e., that you're really
just applying the distributive property.  

So now we have something pretty simple:

  14b = 7

At this point, we might just guess the answer. But we don't have to. 
We can divide both sides of the equation by 14.  And that gives us

  14b    7
  --- = --
   14   14

    b = 1/2

I know that this has been a lot for you to plow through. I'm sorry 
about that, but this is _all_ stuff that you need to know if algebra 
is going to make any sense.  

At this point, you've seen the basic ideas:

  1) Variables are just names we give to blanks that we need
     to fill in with numbers.

  2) To solve an equation, you keep adding, subtracting, 
     multiplying, or dividing both sides by the same quantities
     until you end up with something simple enough to see
     what has to go in the blanks, usually all the way to
     something like

         ____ = 1/2
     (Later you'll see how to use exponents to deal with roots,
     and logarithms to deal with exponents, but it's all the 
     same idea.) 

  3) You use the distributive property to combine 'like terms'
     (i.e., terms that have the same variables, raised to the
     same exponents).  

Once you have these down, the rest is mostly tricks that apply to
particular situations.  

Now take another crack at your original problem, 

  13.7b - 6.5 = -2.3b + 8.3

and show me the steps you take.  This is very similar to the problem
we just went through.  Only the numbers are different.  

- Doctor Ian, The Math Forum 

Date: 05/11/2003 at 13:49:59
From: Vinnie
Subject: Algebra

So it is like this:

1)   13.7b - 6.5 = -2.3b + 8.3
   +         6.5           6.5  ---> You add 6.5 to both sides.
   ---------------------------       This cancels on one side, and   
         13.7b = -2.3b + 14.8        gets added to 8.3 on the other.

2)       13.7b = -2.3b + 14.8       
   +      2.3b    2.3b          ---> You add 2.3b to both sides.
   ---------------------------       This cancels on one side, and
           16b =         14.8        gets added to 13.7b on the other.

3)         16b   14.8               
           --- = ----           ---> You divide both sides by 16.
           16     16                 This cancels on one side,
                                     and on the other you get 0.925.
4)           b = 0.925          ---> Nothing left to do.  
Like that?

Date: 05/11/2003 at 16:14:25
From: Doctor Ian
Subject: Re: Algebra

Hi Vinnie,

Yes, just like that. Is it making more sense now? 

- Doctor Ian, The Math Forum 

Date: 05/11/2003 at 16:23:08
From: Vinnie
Subject: Thank you (Algebra)

Yes. Thanks!
Associated Topics:
Middle School Equations

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