Algebra (Re.: Reiny)
posted by Sara .
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^2a^2)
and I plug in 3x+4 for a and 3x4 for b.
[(3x+4)^2+(3x4)^2]/[(3x4)^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+(3x4)^2]/[(3x4)^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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