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Forget the name of it, but i need a review

  1. Jun 11, 2009 #1
    1. The problem statement, all variables and given/known data

    I basiclly just need a little review and maybe some 1 to show me how to solve out the equation/and equation like it. here is an example:

    2x+1 / x-1 + 1/ x+1 = ?

    2. Relevant equations
    ?



    3. The attempt at a solution
    ?
     
  2. jcsd
  3. Jun 11, 2009 #2

    rock.freak667

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    I assume by ? you mean a number like 3 or 2 or even some function f(x)?


    but here's a clue

    [tex]\frac{1}{a} + \frac{1}{b}= \frac{a+b}{ab}[/tex]
     
  4. Jun 11, 2009 #3
    by Question mark i mean it would equal like 2x + 1 / x^2 - 1 or something along those lines
     
  5. Jun 11, 2009 #4

    rock.freak667

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    then just bring them to the same common denominator (x-1)(x+1). Do you know how to do that?
     
  6. Jun 11, 2009 #5

    HallsofIvy

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    I THINK you mean you want to add the two fractions (2x+1)/(x-1)+ 1/(x+1).
    (Please use parentheses!)

    As Rock.freak667 said, you need to get "common denominators". Here, the common denominator is (x-1)(x+1). Multiply numerator and denominator of the first fraction by x+1 and numerator and denominator of the second fraction by x-1.
     
  7. Jun 11, 2009 #6
    no i forget, do you foil it ?
     
  8. Jun 11, 2009 #7
    yeah, ok I did that and i got

    (2x^2-x-1)/(x2-2x+1) + (x+1)/ (x^2 -2x-2)

    now what
     
  9. Jun 11, 2009 #8

    rock.freak667

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    the denominator would be (x-1)(x+1)

    now do as HallsofIvy said and

     
  10. Jun 11, 2009 #9
    that just confused me. i did what halls said and i got the answer you quoted. so i did it wrong?
     
  11. Jun 11, 2009 #10

    rock.freak667

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    [tex]\frac{2x+1}{x-1} + \frac{1}{x-1}[/tex]



    [tex]= \frac{(2x+1)}{(x-1)(x+1)} + \frac{(x+1)}{(x-1)(x+1)}[/tex]

    since the denominators are both the same, you can just add the numerators now.
     
  12. Jun 11, 2009 #11
    did you copy that down wrong? its

    (2x+1)/(x-1)+ 1/(x+1).
     
  13. Jun 11, 2009 #12

    rock.freak667

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    Sorry there, forgot to do some more typing, should be this



    [tex]
    = \frac{(2x+1)(x+1)}{(x-1)(x+1)} + \frac{(x-1)}{(x-1)(x+1)}
    [/tex]
     
  14. Jun 11, 2009 #13

    HallsofIvy

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    Now, austin1250, use "FOIL" to multiply (x-1)(x+1) carefully! It is NOT [itex]x^2- 2x- 1[/itex].
     
  15. Jun 12, 2009 #14
    oh did i say that? Sorry

    x2-1
     
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