How Variables and Equations Work

(An archive question of the week)

We’re looking at extended discussions of a single topic, which illustrate how we try to guide a student to a deeper understanding. Here, a student asks how to solve an equation, and Doctor Ian takes him through the whole process, clarifying what it means to solve an equation, and what you do to get there.

The problem

It came in 2003, from Vinnie:

Variables, Explained


  13.7b - 6.5 = -2.3b + 8.3

Doctor Ian started by observing that there are many possible reasons for being unable to solve such an equation, so he asked a diagnostic question:

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.)

This is a good first step to find out what a student needs: Try a simpler problem. This will reveal whether Vinnie makes little mistakes, or big ones. Here’s the response:

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?

Aha! There are some fundamental errors here; it looks superficially like what we expect to see, but every single detail is wrong. Ian adeptly avoids saying that, though!

What is a variable?

He starts at the beginning, with what first a variable, then an equation means:

Hi Vinnie,

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

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.

Thinking of a variable as standing for a specific number is important; and putting specific numbers in place of it goes a long way to demonstrate that idea. We’re also getting at Vinnie’s first big error, namely combining unlike terms (the 14b and the 6). Students often hear about “combining like terms” and may learn the rules, but do they always understand why they can’t combine unlike terms?

What does it mean to solve an equation?

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.

So the equation is a sentence, saying that this number is equal to that number. We want to find what value b must have, if the sentence tells the truth.

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!)

And if the answer turned out to be -2.357, we might never get to guessing that! But by “guessing and checking” (trying out the equation with specific values), we are again focusing on what the equation means, which is essential.

Now (still not following rules someone has taught us, but just thinking), we can find a solution by working backward:

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 that

             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.

So, we ask, “What plus 2 is 11?” and answer, “It’s 11 – 2, which is 9.” Then we ask, “What times 3 is 9?” and answer, “It’s 9 divided by 3, which is 3.”

Checking a solution

There’s one more really important thing about what it means to solve an equation: Since solving means finding the value of the variable that makes the equation true, we can check our answer by seeing if it makes the equation true:

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 for.

So, 3 + 3 is 9, and 9 + 2 is 11, which is just what we wanted.

I’m often amazed, both as a Math Doctor and in face-to-face tutoring, how many students don’t know how to check their answers – which means they’ve missed the whole point of solving!

Are we there yet?

Now it’s time to see how Vinnie is doing:

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?

Vinnie’s response shows that he understands at least what it means to solve an equation, and to check it; but he’s not quite there yet:

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.

The idea that x is “nothing” is interesting; he seems to mean that if you just drop the x, then 5 – 4 = 1 is true. And, in fact, setting x to 1 does exactly that, as 5x is just 5.

More on variables

Doctor Ian continued with the concept of variables, to make sure a good foundation had been laid for the next step. (Keep in mind that this is 2003.)

Hi Vinnie,

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

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.

Blanks could be all we need, if we never had more than one unknown:

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.

In computer programming, variables are usually whole words, or even phrases, because there are so many of them to keep track of.

Not-so-nice answers

Doctor Ian next confirmed the correct answers to the first two problems, then dealt with the last, 3x + 5 = 2, using the same “something” approach as before:

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.

That was not really harder technically; but the negative number gave it a different feel. Often students are troubled when the answer to a problem is not as nice as the answers to previous examples, and they think they must be wrong.

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.

Never expect things to be nice! We learn algebraic techniques initially in cases where the algebra seems hardly necessary; but we are preparing for the real world, which is not so nice. When a teacher always gives nice examples, it is a disservice to the students, not a kindness in the long run.

Some essential tools

It’s time now to start heading toward standard methods, in order to make this solving process more routine. For this, Doctor Ian moves on to somewhat harder problems, with the variable on both sides:

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)

A key idea here, which many students miss at first, is that the “something” that we do to both sides must be an operation we perform on the value of each entire side. The two examples here are adding a number to each side, and multiplying each side by a number. We can’t do something like multiplying one term of each side by the same number.

Choosing the appropriate tool

This tells us what we are allowed to do – what “tools” we have available. But the examples above didn’t actually accomplish anything, did they? The other thing we need to know is how to decide what to do – that is, what “tool” is useful at a certain point in the process.

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.

Since the final goal (an equation saying what the value of the variable is: b = 1/2) is simple, anything that makes the equation simpler (but still true) is useful. It’s like trying to find water in the woods by always hiking downhill.

And what made this particular choice (adding 6) useful? It reduced the number of terms in the equation. Note that we aren’t following a set routine yet; this isn’t necessarily “the right” thing to do, or even “the best” – it’s just a useful thing to do, and that’s enough. You don’t need to memorize a routine (though you probably will, eventually, just by doing the same thing many times). The emphasis here is on doing something helpful.

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

What was it that made this useful? We still have the same number of terms, so in a sense it isn’t simpler; but we now have the variable on only one side, which is another characteristic of our goal, which will have only b on the left side.

Another tool: the distributive property

In addition, it sets us up to use another important tool:

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 equation:

  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 subtraction:

      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.

The distributive property says that a multiplication with a sum, \(a(b + c)\), can be rewritten as a sum of multiplications, \(ab + ac\). We just multiply each term in the sum by the multiplier, “distributing” the multiplication. Here, we did the reverse: combine a sum by removing a common factor from each term and multiplying the resulting sum by it: \(ab + ac = a(b + c)\). In the particular situation we have here, where the common factor is the variable, we call the process combining like terms: \(ax + bx = (a + b)x\).

I like to think of what we just did this way: If you have 16 “bees” and take away “2 “bees”, there are 14 “bees”. So it’s just common sense. But it also has those fancy names.

Back to Doctor Ian:

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.

I find that a lot of students, if they decide just to guess, tend to say that b is 2 instead of 1/2. They see the 14 and the 7, and they jump to a conclusion. That’s why it’s useful to explicitly go through this division step, undoing the multiplication by dividing both sides, to make sure everything ends up in the right place. That’s also why you need to check your answer – and checking is all the more important when you are most confident (so that you may have rushed), or when you least want to check (because the answer is a fraction, and you want to just write it down and run).


Doctor Ian summarized what he’d said, and then did a final check:

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.

Vinnie successfully solved the original problem, explaining the work much as Doctor Ian did, but displaying the work as teachers often do:

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.
           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.

And, of course, to finish up, we should check the answer:

Left-hand side: \(13.7b – 6.5 = 13.7(0.925) – 6.5 = 6.1725\)

Right-hand side: \(-2.3b + 8.3 = -2.3(0.925) + 8.3 = 6.1725\)

Since these are equal, the solution is correct.

Seeing a student solve his own problem with no direct help is our goal in tutoring. We seek not to give answers, but understanding. And this whole interaction is typical of the best we do.

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