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Algebraic Problem

  1. Mar 27, 2004 #1
    This question kinda stumped me. Can any1 post the answer with the working and all? Thanks :biggrin:

    Show that

    [tex]\frac{ 3 \left( \frac{x+5}{x-1} \right) + 4 }{ 4 \left( \frac{x+5}{x-1} \right) + 1 }[/tex] = (3x+11)/{5x+19)
     
    Last edited: Mar 28, 2004
  2. jcsd
  3. Mar 27, 2004 #2

    Hurkyl

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    Sorry, we don't do your homework for you.

    Anyways, you seem to be missing parentheses, did you mean:

    [tex]
    \frac{ 3 \left( \frac{x+5}{x-1} \right) + 4 }
    { 4 \left( \frac{x+5}{x-1} \right) + 1 }
    [/tex]

    ?


    Anyways, what have you tried to do to solve this problem?
     
    Last edited: Mar 27, 2004
  4. Mar 27, 2004 #3
    Yes, thats what i meant, but i was unsure on how to write it in that format. Well first, i tried to cross multiply but thats the problem..I'm not sure how to..
     
  5. Mar 27, 2004 #4

    Hurkyl

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    Well, in general, cross multiplication says that the equation

    p/q = r/s

    is equivalent to

    ps = qr (with q and s inequal to 0)


    Are you having trouble seeing how to make this substitution, or is it the next steps?
     
  6. Mar 27, 2004 #5
    yep, it's the substitution thats giving me the problem so far

    4((x+5/(x-1))+1 * 3x+11 only this one though
     
    Last edited: Mar 27, 2004
  7. Mar 27, 2004 #6

    Hurkyl

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    You have to put [ tex ] and [ /tex ] tags (no spaces) around the LaTeX code.


    You're missing the ) after x + 5, but I think that's just a typo.

    The big thing that you might be doing wrong is that you didn't put parentheses around each of the terms there. What you want is

    [tex]
    ( 4 ( \frac{x+5}{x-1} ) + 1 ) (3x + 11)
    [/tex]
     
  8. Mar 27, 2004 #7
    [tex](4 ( \frac{x+5}{x-1} + 1 ) (3x+11)[/tex]

    hmm ok i think i got the hang of it :cool:
    so can you tell me exactly how do i go about multiplying these two terms?
     
    Last edited: Mar 27, 2004
  9. Mar 27, 2004 #8

    Hurkyl

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    You're missing a parenthesis again!


    There are at least two ways to progress from here.

    One way is to look at the equation as a whole imagine the complicated thing is replaced by a simple thing; do you know how to expand (4z+1)(3x+11)?

    The other way is to look at little pieces. Do you know any way to combine 4 (x+5)/(x-1) + 1 into one term?
     
  10. Mar 27, 2004 #9
    hmmm i expanded (4x+1)(3x+11) and got 12x2+47x+11

    I'm unsure about the other method
     
  11. Mar 28, 2004 #10

    Hurkyl

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    Try the same procedure, but on (4z+1)(3x+11) instead of (4x+1)(3x+11).

    (We are justified in making a new letter to represent the fraction (x+5)/(x-1), but it has to be a new letter; replacing the fraction with x won't work)
     
  12. Mar 28, 2004 #11
    Oh ok, i thought that was a typo :rolleyes:

    Ok i got 12xz+44z+3x+11
     
  13. Mar 28, 2004 #12

    Hurkyl

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    That looks right. Now, since z was a substitute for the fraction (x+5)/(x-1), if you substitute the fraction back in for z, you will have successfully multiplied the two terms you had trouble with!
     
  14. Mar 28, 2004 #13
    Ohhhhh ok i see :biggrin: Lol i didn't think of it that way. However I am still in a bit of a jam :confused: So i'm left with:

    15x-5({x+5}/{x+1}) = 12x({x+5}/{x-1}) + 44({x+5}{x-1}) + 3x + 1

    Or something like that :frown:
     
    Last edited: Mar 28, 2004
  15. Mar 28, 2004 #14

    Hurkyl

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    Well, there are (at least) again two approaches. :)

    (a) Combine each side into a single fraction
    (b) Clear the denominators (by multiplying both sides by the least common denominator)


    And I'm off to bed.
     
  16. Mar 28, 2004 #15
    hmmmm, my previous equation was incorrect. Here is the correct one:

    57(x+5)/(x-1) + 15x(x+5)/(x-1) +20x + 76 = 12x(x+5)/(x-1) + 44(x+5)/(x-1) + 3x + 11

    Then i subtract then get:

    [tex]13(\frac{x+5}{x-1}) + 3x(\frac{x+5}{x-1}) = -17x -65[/tex]

    =[tex]\frac{13x(x+5) +3x(x+5)}{x+1}[/tex]


    Any errors? :redface:
     
    Last edited: Mar 28, 2004
  17. Mar 28, 2004 #16

    Hurkyl

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    13x doesn't look right
     
  18. Mar 28, 2004 #17
    [tex]\frac{13(x+5) +3x(x+5)}{x-1}[/tex]

    Oops sorry thats what i meant! OHHHHHHHHH YAY!! Thanks a lot ,i'm finally seeing the answer :biggrin: :biggrin: :biggrin: I got x is either equal to 0 or -3.8 :smile:
     
    Last edited: Mar 28, 2004
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