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OfflineGrav
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Registered: 02/06/02
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algebra question
    #2090507 - 11/10/03 07:40 PM (20 years, 4 months ago)

ok this fuckin problem is driving me nuts...

can anyone show me step by step how to simplify this complex fraction?

........3
3 + ----
........X
_________
........3
3 - ----
........X


!??!?!

the book says the answer is: "x squared over (x - 1) squared", but i cant see how to get there.

Edited by Grav (11/10/03 07:44 PM)

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Invisibleoneducktwoducks
Registered: 12/13/02
Posts: 2,321
Re: algebra question [Re: Grav]
    #2090519 - 11/10/03 07:43 PM (20 years, 4 months ago)

Are the fractions like [ (3 + 3) / x ] or [ (3 / x) + 3 ]???

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OfflineGrav
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Re: algebra question [Re: oneducktwoducks]
    #2090522 - 11/10/03 07:45 PM (20 years, 4 months ago)

i was havin trouble typing it out right...

(3 / x) + 3

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OfflineGrav
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Re: algebra question [Re: Grav]
    #2090530 - 11/10/03 07:48 PM (20 years, 4 months ago)

i get

3x^2
_________
3x^2 - 6x - 3


does that simplify to

x^2
_____
(x - 1)^2

??

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Offlinekosmic_charlie
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Re: algebra question [Re: Grav]
    #2090539 - 11/10/03 07:51 PM (20 years, 4 months ago)

Sorry dude I tried but that is a difficult problem. Maybe I'll keep trying.


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Invisibleoneducktwoducks
Registered: 12/13/02
Posts: 2,321
Re: algebra question [Re: Grav]
    #2090563 - 11/10/03 08:06 PM (20 years, 4 months ago)

I just got off the phone, so here is my solution. Just let me know if you don't understand any of my notation.

I made common denominators in the numerator and denominator to get:

[ (3x + 3) / x ] / [ (3x - 3) / x ]

Then you multiply by the reciprocal of the denominator, so:

[ (3x + 3) / x ] * [ x / (3x - 3) ] = (3x + 3) / (3x - 3)

And that simplifies to (x + 1) / (x-1)

I can't remember if I'm performing any illegal operations though, so let me know if you see any holes with my method.

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OfflineGrav
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Re: algebra question [Re: oneducktwoducks]
    #2090639 - 11/10/03 08:34 PM (20 years, 4 months ago)

hrmm... well for the common denominator for the first term wouldnt it have to be (x - 1) ?

i figured it out though trying it your way ... it came out the same answer as it did my method but i noticed the -3 at the end of the denominator was supposed to be a +3 which let me factor the denominator into (x-1)^2

sweet


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Invisibleoneducktwoducks
Registered: 12/13/02
Posts: 2,321
Re: algebra question [Re: Grav]
    #2090656 - 11/10/03 08:43 PM (20 years, 4 months ago)

If you just take the expression:

(3 / x) + (3 / 1)

The least common denominator will be x, so you have:

(3 / x) + (3x / x)

Then you simplify and get:

[ (3x + 3) / x ]

Am I missing something? I don't see where the (x - 1) your talking about could be.

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OfflineGrav
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Re: algebra question [Re: oneducktwoducks]
    #2092601 - 11/11/03 01:16 PM (20 years, 4 months ago)

shit, my bad

the top fraction had a (x-1) denominator, not x

what math level are you at, Merletto?

im starting computer science in school, unfortunately i dont know jack shit with my math yet

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Invisibleoneducktwoducks
Registered: 12/13/02
Posts: 2,321
Re: algebra question [Re: Grav]
    #2092620 - 11/11/03 01:20 PM (20 years, 4 months ago)

Calc 1, but I think I know earlier maths pretty well (780 on SAT math section, 760 on SAT2 for math). In fact, I have math class in 15 minutes =(

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OfflineIts Pat
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Re: algebra question [Re: Grav]
    #2093741 - 11/11/03 06:43 PM (20 years, 4 months ago)

Please re-write your problem again now that your know how to write it!


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Invisibleoneducktwoducks
Registered: 12/13/02
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Re: algebra question [Re: Its Pat]
    #2093804 - 11/11/03 07:01 PM (20 years, 4 months ago)

I think he means it's:

[ 3 + (3 / {x - 1}) ] / [ 3 - (3 / x) ]

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OfflineAnnom
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Re: algebra question [Re: Grav]
    #2095682 - 11/12/03 03:56 AM (20 years, 4 months ago)

That wasn't easy....:



Last step:

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