Friday
May 6, 2016

# Homework Help: Algebra (Re.: Reiny)

Posted by Sara on Friday, July 20, 2007 at 5:44pm.

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

so, I got that,and it's clear now. Here, I'll show you how I am puttin in the values and how it doesn't add up for me. Could you tell me then what i am still doing wrong?

O.k.
so I have (a^2+b^2)/(b^2-a^2)
and I plug in 3x+4 for a and 3x-4 for b.
[(3x+4)^2+(3x-4)^2]/[(3x-4)^2-(3x+4)^2]

So here is my next question. Can I get rid of something here or do I need to solve each term, like (3x+4)(3x+4) [for (3x+4)^2] and get 9x^2+12x+12x+16 ...then I continued this strategy but I just don't get the answer. Can you solve the rest as well and explain what you are doing?
Your help is greatly appreciated!!

I'll take it from

[(3x+4)^2+(3x-4)^2]/[(3x-4)^2-(3x+4)^2]

let's just expand this:
(9x^2 + 24x + 16 + 9x^2 - 24x + 16) / (9x^2 - 24x + 16 -(9x^2 + 24x + 16))

= (18x^2 + 32)/(-48x) divide each term by 2
=(9x^2 + 16)/(-24x)

thank you sooo much!
At some point in my notes I got to (18x^2 + 32)as well, but what I didn't know and what caused my confusion is that the - sign in front of 9x^2 functions like a negative sign in front of brackets changing the signs within...I didn't use brackets when I tried it and so only the 9's cancled out. Please take a look at my other postings. You've been a great help!!
Thank you!

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