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Solving a Multi-Variable Formula or Equation for One Variable

Date: 07/24/2008 at 02:23:27
From: Tori
Subject: Solving an equation with multiple variables

I was given y^2 + 3yz - 8z - 4x = 0 and asked to solve for z.    

I haven't the slightest idea as to how to get rid of the y in 3yz.  It
seems like if you do manage to get rid of it, you have to put it back
again later.

I know you can factor out a y from the first half, and a -4 from the
other half so it looks like this:  y(y + 3z) - 4(2z + x) = 0.  After
that, I'm stuck.



Date: 07/24/2008 at 16:54:39
From: Doctor Ian
Subject: Re: Solving an equation with multiple variables

Hi Tori,

Suppose we start with something like

  xz + yz = 1

and we want to solve for z.  We can use the distributive property to
factor z out of the expression on the left, and then divide:

  z(x + y) = 1

         z = 1/(x + y)

Can you see how to apply this same reasoning to your problem? 

- Doctor Ian, The Math Forum
  http://mathforum.org/dr.math/ 



Date: 07/24/2008 at 23:44:08
From: Tori
Subject: Solving an equation with multiple variables

Here's what I did:

  y^2 + 3yz - 8z - 4x = 0
  y(y + 3z) - 8z - 4x = 0
       y(y + 3z) - 8z = 4x
          y + 3z - 8z = 4x/y
                  -5z = (4x - y)/y
                   5z = (-4x + y)/y
                   5z = -4x
                    z = (-4/5)x



Date: 07/25/2008 at 08:55:09
From: Doctor Ian
Subject: Re: Solving an equation with multiple variables

Hi Tori,

I'm glad you wrote back!  It looks like there are a couple of things
that we need to clear up.  You wrote:

>Here's what I did:
>
>  y^2 + 3yz - 8z - 4x = 0
>  y(y + 3z) - 8z - 4x = 0
>       y(y + 3z) - 8z = 4x
>          y + 3z - 8z = 4x/y

You ran into a problem here.  Note that, when you divide an equation
on both sides by y, you have to divide EVERYTHING on both sides by y,
and you didn't do that.  Here's how that step should have looked:

    y(y + 3z) - 8z = 4x
       
   (y + 3z) - 8z/y = 4x/y

See the difference?  

>                 -5z = (4x - y)/y
>                  5z = (-4x + y)/y
>                  5z = -4x

You did a similar thing here...

So, it looks to me like there are a couple of things going on here. 
The first is that you're not quite clear on the idea of cancellation.
That is, you're doing things like this,

  -4x + y
  -------  ->  -4x
     y

with variables, that you would never do with constants.  For example,
it would never occur to you to do this,

  5 + 2
  -----  ->  5
    2

but that's essentially what you did, isn't it? 

The idea of cancellation is basically the same as the idea of
equivalent fractions.  That is, we can start with a fraction like 2/3,
and multiply by 5/5, and get a fraction with the same value:

  2   5   10
  - * - = --
  3   5   15

Similarly, if we start with 10/15, we can break it into factors, and
cancel some of those by separating them out:
 
  10   2 * 5   2   5   2       2
  -- = ----- = - * - = - * 1 = -
  15   3 * 5   3   5   3       3

But this only works with MULTIPLICATION.  If you try it with addition,
you get nonsense:

  10    7 + 3        7
  -- = ------ = ... --        Wrong!
  15   12 + 3       12


  21   9 + 12
  -- = ------ = ... = 9       Wrong!
  12     12

And if we can't do something with numbers, we can't do it with
variables--because variables are just numbers whose values we don't
happen to know at the moment.  They still have to obey all the same
rules. 

This, by the way, is one of the reasons we spend so much time learning
to factor polynomials.  We can't do this,

  x^2 + 3x + 2      3x + 2
  ------------  ->  ------      Wrong!
  x^2 + 6x + 5      6x + 5

but we CAN do this,

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

Note that in this case, we have to stick a caveat on our final expression:

  x^2 + 3x + 2   x + 2
  ------------ = -----      Where x is not -1
  x^2 + 6x + 5   x + 5

This is because setting x to -1 in the ORIGINAL expression would cause
a division by zero, which is undefined. 

Anyway, does this make sense?  
  
*****

The second thing is that it looks like you sort of lost sight of the
goal early on.  You want to solve for z, so you want to isolate the
quantities involving z early on.  For example, 

  y^2 + 3yz - 8z - 4x = 0

       y^2 + 3yz - 8z = 4x

             3yz - 8z = 4x - y^2

Now we can factor out z, 

            z(3y - 8) = 4x - y^2

and use division to cancel it on the left, 

                    z = (4x - y^2)/(3y - 8)

Note how I'm using ()'s to make sure everything that needs to be
divided, gets divided!

My point is, factoring out a y, 

   y^2 + 3yz - 8z - 4x = 0

   y(y + 3z) - 8z - 4x = 0

isn't incorrect, but it's not especially helpful, either.  

Since you are trying to isolate z, the key is to see your original
equation as
              
  [stuff] + [stuff]z - [stuff]z - [stuff] = 0

You don't really care what that other stuff is.  It's just a
distraction.  Put any of the z terms on one side, and the non-z terms
on the other.  Factor the z out of the z terms, then divide both sides
by the other factor to isolate the z.

Does this make sense? 

- Doctor Ian, The Math Forum
  http://mathforum.org/dr.math/ 



Date: 07/26/2008 at 02:40:31
From: Tori
Subject: Thank you (Solving an equation with multiple variables)

Ah!  That makes a lot more sense.  I always seem to forget what you
can and can't cancel out in fractions.  I totally didn't see that you
could factor out z's, too.  Thank you for your help!
Associated Topics:
High School Polynomials

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