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Express sum as single algebraic fraction

  1. May 4, 2007 #1
    1. The problem statement, all variables and given/known data

    Express:

    [tex]\frac{1}{x-2}+\frac{2}{x+4}[/tex]


    3. The attempt at a solution

    Well I got x² + 2x - 11 = 0

    but I think that is wrong
     
  2. jcsd
  3. May 4, 2007 #2

    Integral

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    How did you get your solution? Show us!
     
  4. May 4, 2007 #3
    1 + 2 = (x-2)(x+4)
    3 = x² + 2x - 8
    then take 3 from both sides gives you the answer I previsouly posted... bust that doesn't seem right as how do I kow

    [ex]\frac{1}{x-2}+\frac{2}{x+4}[/tex] = 1

    is equal to 1
     
  5. May 4, 2007 #4

    Integral

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    To clear fractions you must cross multiply.

    [tex] \frac a b + \frac c d = \frac {ad + bc} {bd} [/tex]

    is that how you did it?
     
  6. May 4, 2007 #5
    It shouldn't equal anything :-/
     
  7. May 4, 2007 #6

    VietDao29

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    Nope, the first line is just sooooo wrong, you cannot do that. :frown:
    What you should do is to make common denominator, or in other words, cross multiply, as Integral has pointed out:
    [tex]\frac{a}{b} + \frac{c}{d} = \frac{ad}{bd} + \frac{bc}{bd} = \frac{ad + bc}{bd}[/tex]
    Ok, I'll give you an example:
    [tex]\frac{1}{x - 5} + \frac{2}{x} = \frac{x + 2 (x - 5)}{x (x - 5)} = \frac{3x - 10}{x ^ 2 - 5x}[/tex].
    Can you get it? :)
     
  8. May 5, 2007 #7
    [tex]\frac{3x}{x^{2}+2x-8}[/tex]

    is what I got
     
  9. May 5, 2007 #8

    Hootenanny

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    You might want to try that again. Here, I'll start for you;

    [tex]\frac{1}{x-2}+\frac{2}{x+4} = \frac{1(x+4)+2(x-2)}{(x-2)(x+4)}[/tex]

    Can you simplify that any?
     
  10. May 5, 2007 #9
    [tex]\frac{x+4+2x-4}{x^{2}+2x-8}[/tex]

    which is hen simplified to

    [tex]\frac{3x}{x^{2}+2x-8}[/tex]
     
  11. May 5, 2007 #10

    Hootenanny

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    Sorry, my bad. I had a sign error, you are of course correct!
     
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