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Eliminatory Excess?

Date: 08/31/2015 at 11:03:26
From: cortney
Subject: solving stytems of equations three vaitables  (elemination)

I'm supposed to eliminate one variable from this system of equations:

    -x - 5y - 5z = 2
   -4x - 5y -  z = -20
     x + 5y - 2z = -20

I'm trying to get rid of x in the first and second equations, but the y is 
not letting me do that. The x's and y's cancel each other out as soon as 
you put the problem together!

I know how to eliminate variables. I already know that you have to 
multiply the whole equation. But I find this system confusing because in 
the first and last equations, the x and y cancel each other out.

I have seen this type of problem several times before and I never know 
what to do about it.



Date: 08/31/2015 at 11:44:44
From: Doctor Peterson
Subject: Re: solving stytems of equations three vaitables  (elemination)

Hi, Cortney.

When something unexpected happens, you can either change your strategy to 
avoid it (e.g., eliminate a different variable or use a different pair of 
equations), or think of it as enabling you to skip a step. It's sort of 
like when you are going upstairs and think there is another step, so if 
you are not paying attention, you can stumble when your foot touches 
nothing. If you ARE paying attention, you just conclude, "I got there 
sooner than I expected!"

So let's take what you were doing, and see how it is not really a problem. 

You have this system:

    -x - 5y - 5z =   2  [A]
   -4x - 5y -  z = -20  [B]
     x + 5y - 2z = -20  [C]

You chose to add equations [A] and [C], intending to eliminate x. But you 
found that you accidentally eliminated both x and y:

   [A]   -x - 5y - 5z =   2
   [C]    x + 5y - 2z = -20
        -------------------
                  -7z = -18

All that happened is that you prematurely solved for a variable:

   z = 18/7

That's a GOOD thing! Now you have to find x and y, so you'll have to do 
something a little different -- but not that different: just continue your 
plan and eliminate x from a different pair of equations, say [B] and [C]:

    [B]  -4x -  5y -  z =  -20
   4[C]   4x + 20y - 8z =  -80
         ---------------------
               15y - 9z = -100

Your original plan at this point was to have a pair of equations in the 
two variables y and z. It's actually a little better than you planned: you 
have already solved for z, and you have another equation that you can 
solve for y by substituting 18/7 for z. Then you can carry on with the 
standard plan by substituting for both y and z in any of the original 
equations, and solving for x.

(By the way, the fractions we're getting suggest an error in what you 
wrote. It may be that the solution really does involve fractions, but 
double-check the original statement of the problem to make sure.)

The standard plan is to combine pairs of equations to get two equations in 
two unknowns, combine those to solve for one variable, then go back and 
pick up the variables you'd eliminated. I think of this procedure as like 
climbing a mountain, then going back down and picking up the equipment you 
discarded on the way.

          [F]           one variable
         /   \
      [D]     [E]       two variables
     /   \   /   \
   [A]    [C]     [B]   three variables

Your equation [D] happens to have only one variable, but you can still use 
it just as planned.

If finding something unexpected throws you for a loop, you can also change 
your plan, as I said earlier, and choose a different strategy. You might 
avoid eliminating x or y first, and instead eliminate z by combining [A] 
and [B], then [B] and [C]. This will work fine.

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



Date: 08/31/2015 at 12:01:21
From: cortney
Subject: Thank you (solving stytems of equations three vaitables  (elemination))

OMG thank you so much. I never realized that this could even happen!
Associated Topics:
High School Linear Equations

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