# Algebra

Posted by Tammy on Thursday, July 19, 2007 at 10:12pm.
(3x+4)/(3x-4)+(3x-4)/(3x+4)
--------------------------- Divide top by bottom
(3x-4)/(3x+4)-(3x+4)/(3x-4)

Very Big Problem and help would be greatly appreciated

Algebra - Reiny, Thursday, July 19, 2007 at 11:59pm
not that tough once you notice that your question has the pattern
(a/b + b/a)÷(b/a - a/b)
where a = 3x+4 and b=3x-4

this easily reduces to (a^2 + b^2)/(b^2 - a^2)

now replace a and b with 3x+4 and 3x-4 and it simplifies to

(9x^2+16)/(-24x)

hmm, the answer is right (it's in my book) but I don't understand how you simply reduce it to (a^2+b^2) (b^2-a^2)
and from there when I tried to replace a and b with 3x+4 and 3x-4 I don't know how to simplify it or get the answer. Could you write down the steps detailed? I have a bunch of problems that I can't solve that are similar, so I think if I understand the concept of this one I will be able to sove all the other ones.

The fact that there was that nice symmetry to the question, makes it easier than it first appears.

I had it as:

(a/b + b/a)÷(b/a - a/b)

look at (a/b + b/a)

the common denominator for this would be ab, so
a/b + b/a = (a^2 + b^2)/ab

similarly for

(b/a - a/b) the common denominator is ab and this reduces to
(b^2 - a^2)/ab

so now we have
(a^2 + b^2)/ab ÷ (b^2 - a^2)/ab

remember when dividing one fraction by another, we multiply by the reciprocal of the second fraction, so

(a^2 + b^2)/ab ÷ (b^2 - a^2)/ab

=(a^2 + b^2)/ab * ab/((b^2 - a^2)

the ab's cancel and that's how I got

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

Now replace the original values of a and b and you should be able to do that yourself.

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