Algebra (Re.: Reiny)

posted by .

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?

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

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


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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