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Clearing Fractions in an Equation

Date: 01/13/2005 at 01:24:10
From: Karen
Subject: it's about algebra; clearing an equation of fractions,
decim

I have this math problem that I understand how to do but it's just
that there's this one that I'm really stuck on.  I get confused
because it has parentheses and the way my teacher explains it we need
to use more parentheses.  Here is the problem:

1/4(8y + 4) - 17 = -1/2(4y - 8)

Here's what I tried:

1/4(8y + 4) - 17 = -1/2(4y - 8)
4[1/4(8y + 4) - 17] = 4[-1/2(4y - 8)]
4(1/4)4[(8y) + 4(4) - 4(17) = 4[(-1/2)4(4y) - 4(8)

This was the most I could do.  I hope I'm doing it right because I was
really struggling.  On other problems I don't get as confused as I did
on this one.



Date: 01/13/2005 at 02:09:03
From: Doctor Ian
Subject: Re:  it's about algebra; clearing an equation of fractions,
decim

Hi Karen,

To be honest, in this case I'd go ahead and distribute the fractions
before clearing them, because that actually _will_ clear them:

     1/4(8y + 4) - 17 = -1/2(4y - 8)

  (8y/4 + 4/4) - 17 = (-4y/2 + 8/2)

      (2y + 1) - 17 = -2y + 4

Look, no fractions.  :^D

It's easy to fall into the trap of trying to classify situations in
terms of rules, e.g., "There are fractions, so I need to multiply both
sides of the equation by the common denominator to clear them".  It's
true that often, that's the best way to go.  But there are times--
like this one--when it's not.  The key thing to notice here is that
it's easy to take 1/4 of each of the numbers in the left hand binomial
and also easy to take 1/2 of each of the numbers in the right hand
binomial.

Your job is to stay flexible, consider all your options, and choose
the one that is going to make your life easiest in whatever situation
you happen to be in. 

If you decide that you really want to multiply both sides by 4, there
is one key idea that you seem to be missing in your work, which is
making it much harder than it needs to be.

Suppose you have two numbers that are multiplied together, like 2(3).
 We know that's 6, right?  So now if you want to multiply that by 4,
the answer should be 24 since 4*6 is 24.  Let's show the
multiplication by 4 this way:

  4[2(3)] which is the same as 4[6] or 24

The mistake you are making in your work is that you are multiplying
the 4 times the 2 and times the 3, so you are getting:

  4(2)4(3)

But that makes 8*12, which is 96, not the 24 it should be.  You've
multiplied by an extra 4!

The important rule is that when you multiply through an equation to
clear a fraction or decimal, you need to be sure to multiply each
_term_ of the equation by that number.  The terms are separated by
addition and subtraction.  Any group of items that are being
multiplied together are all part of one term.  So, in your equation,
there are really only two terms on the left and one on the right side:

  1/4(8y + 4) - 17 = -1/2(4y - 8)
  -----------   --   ------------
    1 term    1 term    1 term

You might have been told that in a binomial, like (8y + 4), there are
two terms, and that's correct.  If we distribute the 1/4 like we did
above, that 1/4 multiplies against each term in the binomial.  But if
we step back another level and look at the equation, that whole first
piece is just one term because the 1/4 and the binomial are being
multiplied together, jus as with 2(3) in the example above.

So when we multiply by 4, we really only need to multiply each of
those three terms by 4 once:

  (4)(1/4)(8y + 4) - (4)(17) = (4)(-1/2)(4y - 8)
     -------------      ----      --------------

Now, just as with 4(2)(3) above, we only use the 4 once.  So on the
left the (4)(1/4) cancels and just leaves (1)(8y + 4).  On the right,
the (4)(-1/2) makes -2 left to multiply against the (4y - 8):

  (1)(8y + 4) - (4)(17) = (-2)(4y - 8)

Distributing the (1) and the (-2):

  8y + 4 - 68 = -8y + 16

Here's a general look at what went on, replacing the binomials with
"this" and "that" to make it clearer how multiplying each side by 4 works:

  1/4*this - 17 = -1/2*that

Now you can go ahead and clear the fractions,

  4(1/4*this - 17) = 4(-1/2*that)

       this - 4*17 = -2*that

         this - 68 = -2*that

Replacing "this" and "that" with the original binomials, we have:

  (8y + 4) - 68 = -2(4y - 8)
  
Does this help?

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

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