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Fraction Problems

Date: 25 Apr 1995 17:23:39 -0400
From: Sandip Mody
Subject: Fractions and Fractional Equations

Hello, I was having trouble with the following problems:

4/x-3 + 6/x+3 = 2/3

and solving for x  F= w/x+1

The other problem was a^2 + b^2/a^2-b^2 / a-b/a+b - a+b/a-b

Thank You for your help.

Date: 26 Apr 1995 11:15:09 -0400
From: Dr. Ethan
Subject: Re: Fractions and Fractional Equations

        Good questions.  I will assume that you know how to work with
fractions when there are just numbers.  If so, then this will be easy 
because here we do the same thing.  What I will do is work the first and 
last problem; then I will leave the second problem for you to think about. 


                 The first problem is:

        4/x-3 + 6/x+3 = 2/3

Well the thing that we need to do is find a common denominator for the 
first two fractions.  That will be (x-3)(x+3), so we need to multiply the 
first fraction by (x+3)/(x+3) and the second by (x-3)/(x-3).  This will not 
change the value of the equation because both of these reduce to one.  Now 
we have:

        4(x+3)/(x+3)(x-3) + 6(x-3)/(x+3)(x-3) = 2/3

which becomes via addition:

        (4(x+3) + 6(x-3))/(x^2 - 9) = 2/3

Now, we multiply both sides by (x^2 - 9) to get:

        4(x+3) + 6(x-3) = 2(x^2 - 9)/3

Now simplify the left hand side and we have:

        10x - 6 = 2(x^2 - 9)/3

Now multiply both sides by 3 and we have:

        30x - 18 = 2x^2 - 18

the 18's cancel and we are left with:

        30x = 2x^2

Divide by two on both sides:

        15x = x^2

So both 0 and 15 are solutions to this problem.  Yea!

Now to your last problem. Here it is.  I am adding () where I think that 
you meant to put them.  If I am wrong then write back and I will rework 
the problem.

        ((a^2 + b^2)/(a^2 - b^2))/((a-b)/(a+b) - (a+b)/(a-b))

Let's work with the bottom first.  Since we have subtraction we need to find 
a common denominator.  Again it will be (a+b)(a-b), so the first term on the
bottom needs to be multiplied by (a-b)/(a-b) and the second by (a+b)/(a+b) 
so it looks like this:

((a^2 + b^2)/(a^2 - b^2))/((a-b)(a-b)/(a-b)(a+b) - (a+b)(a+b)/(a+b)(a-b))

this adds together to be:

((a^2 + b^2)/(a^2 - b^2))/(((a-b)^2 - (a+b)^2)/a^2 - b^2

Now square the subtracted terms and combine them and we have:

((a^2 + b^2)/(a^2 - b^2))/(-4ab/(a^2-b^2))

Now we can consider the whole fraction, and we can do the old invert and
multiply trick for dividing fractions so the division problem above looks
like this now:

   (a^2 + b^2)     a^2 - b^2
  ------------- * -----------
    a^2 - b^2        -4ab

Well the a^2-b^2 terms cancel and we are left with:

                 -(a^2 + b^2)/4ab

as our answer.  

        I hope that I haven't made a careless mistake.  I have tried to
explain each step.  If you need more explanation or you think that I have
made a mistake, then you can write back.  Also, if you need help on that 
other problem or have other questions please write to Dr. Math.

Ethan, Doctor On Call
Associated Topics:
Middle School Fractions

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