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Substituting Parameters for Numbers


Date: 11/29/2001 at 12:54:30
From: Don
Subject: Equation help

Dear Dr. Math,

Please help me with this. I am at a point where I do not know what 
to do with both sides of the equation

17.20X / 100 = X - 579.59

I have changed to 

(17.20X / 100) + 579.59 = X

When the known answer on both sides will be 699.99, what will I do to 
make the equation = 699.99 ?


Date: 11/29/2001 at 13:25:16
From: Doctor Ian
Subject: Re: Equation help

Hi Don,

Note that

  (ab)/c = a(b/c) = b(a/c)

so

  (17.20X)/100 = (17.20/100)X

Once you can separate the X out, you can use the distributive property 
to combine it with the X on the other side of the equation:

         17.20X / 100 = X - 579.59

         (17.20/100)X = X - 579.59

            (0.1720)X = X - 579.59

   (0.1720)X + 579.59 = X

               579.99 = X - (0.1720)X

               579.99 = X(1 - 0.1720)

Now, I just made a mistake there - I changed 579.59 to 579.99, and 
didn't notice it until I compared my final answer to yours.  Which 
brings up an interesting point.  

In a case like this, instead of writing the constants (0.1720 and 
579.59) over and over again, I would substitute parameters early on, 
work with the parameters, and then un-substitute at the end:

             17.20X / 100 = X - 579.59

             (17.20/100)X = X - 579.59

                       aX = X - b            a = 0.1720
                                             b = 579.59
                   aX + b = X                     |
                                                  |
                        b = X - aX                |
                                                  |
                        b = X(1 - a)              |
                                                  |
                b/(1 - a) = X                     |
                                                  |
    579.59 / (1 - 0.1720) = X      <--------------+


Not only does this save you time (it's much quicker to write 'b' 
instead of '579.59'), but it can help you avoid making errors (a 
change from 'b' to anything else is much less likely, and more 
noticeable, than a change from '579.59' to '579.99'). 

It's often (although not always) true that the fewer symbols you have 
in an equation, the easier it is to see patterns that you can use to 
move to the next step in solving it.  

Does this help?

- Doctor Ian, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Basic Algebra
High School Linear Equations
Middle School Algebra
Middle School Equations

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