Last one (I promise)!!!

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JLorenzo

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Apr 19, 2006
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9/x²-9 + 2x/x-3

I am supposed to perform the indicated operation. I first factor which gives me:

9/(x-3)(x+3) + 2x/(x-3)

Then I have to distribute an (x-3) to 2x

That gives me

9/(x-3)(x+3) + 2x(x-3)/(x-3)(x+3)

My final answer would then be:

2x+9/(x-3)(x+3)

Right?
 
NO. Try \(\displaystyle \L
\frac{{2x^2 + 6x + 9}}{{(x - 3)(x + 3)}}\).
 
To PKA...

That is what I got the first time and like a dope I tried factoring the top. But I cannot seem to get it any lower.
 
TK...

It looks like what PKA said but that can't be factored, I don't think.
 
No, I'm not referring to the final result, only to this very bad step in your process. You did not multiply numerator and denominator by the same thing.

2/2 ==> 2(2)/2(3) = 4/6 <== Doesn't work.
2/2 ==> 2(2)/2(2) = 4/4 <== Does work.
 
Methinks you are missing the point.

From this: 2x/(x-3)

This 2x(x-3)/(x-3)(x+3) is very bad.

This 2x(x+3)/(x-3)(x+3) is Perfect.
 
TK...

You are very patient and I appreciate it. Your thoughts are correct.

So the (x+3)'s would cancel out? And what I would be left with is 2x/(x-3)?
 
Cancel? Why? You just put them there so you would have a common denominator. Don't throw them away.
 
TK...

Perhaps the confusion comes from looking at this for many weeks or I need a stronger prescription of contacts to see what you are doing. I understand the factoring and have gotten dinged on it before. Perhaps if I could see the final answer and work backwards...that may help.
 
Poor Josh...DO NOT give up.

Your expression is: 9/x²-9 + 2x/x-3

By the way, that requires brackets: 9 / (x²-9) + 2x / (x-3)
Code: 
       9            2x
  ------------  +  -----
  (x+3)(x-3)       (x-3)

Now for both fractions to have the SAME denominator:

       9             2x(x+3)
  ------------  +  ----------
  (x-3)(x+3)       (x-3)(x+3)

OK? Now that can be rewritten this way:

   9 + 2x(x+3)
  ------------  
   (x-3)(x+3)

Then doing the multiplication in numerator:

   9 + 2x^2 + 6x
  --------------- 
     (x-3)(x+3)

And finally rearranging numerator as per numerator guru's:

   2x^2 + 6x + 9
  ---------------  
     (x-3)(x+3)

...and that's it  !
 
I GOT IT!!!

Dennis, you explained it wonderfully. Do you want to know what my problem was? I was trying to further factor the numerator and I don't think you can!
 
Re: I GOT IT!!!

JLorenzo said:
...I was trying to further factor the numerator and I don't think you can!
Click to expand...
...that is CORRECT thinking...amazing for a writer :lol:
 
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