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Finding x with fractions

  1. Aug 30, 2012 #1

    wrx

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    1. The problem statement, all variables and given/known data
    1)
    ____1___ __ ___2___ __ ____3____ =0
    3x-7 5x-5 3x+1

    here's the other way of presenting the problem.

    1/(3x-7) - 2/(5x-5) - 3/(3x+1) =0

    ans: x=2/3, 3

    2) _____2_____ + _____1_____ __ _____1_____ =0
    (x^2-36)^1/2 (x+6)^1/2 (x-6)^1/2




    2/sqrt(x^2-36) + 1/sqrt(x+6) - 1/sqrt(x-6) =0

    ans:x=10


    3) (x+7)^1/2 - (x+2)^1/2 = (x-1)^1/2 - (x-2)^1/2

    ans:x=2

    2. Relevant equations



    3. The attempt at a solution

    on problem #3, i tried squaring both side of the equations to cancel out those sqrt root but im thinking im doing it the wrong way because the solution is way too long.
     
    Last edited: Aug 30, 2012
  2. jcsd
  3. Aug 30, 2012 #2

    Simon Bridge

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    Or, possibly:[tex]\frac{1}{3x-7}-\frac{2}{5x-5}-\frac{3}{3x+1}=0[/tex]... have you tried #1 and #2 at all?
    #3 was:[tex] (x+7)^{1/2} - (x+2)^{1/2} = (x-1)^{1/2}-(x-2)^{1/2}[/tex]... notice that x ≥ 2 for the RHS to be real?
     
  4. Aug 30, 2012 #3

    ehild

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    It becomes easier if you rearrange the equation before squaring:

    [tex](x+7)^{1/2} - (x-1)^{1/2}= (x+2)^{1/2} -(x-2)^{1/2}[/tex]

    ehild
     
  5. Aug 31, 2012 #4

    ehild

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    Taking the square of both sides and expand

    [tex](x+7) + (x-1)-2\sqrt{(x+7)(x-1)}= (x+2) +(x-2)-2\sqrt{(x-2)(x+2)}[/tex]
    the x terms cancel, simplify and divide by 2:
    [tex]3-\sqrt{(x+7)(x-1)}= -\sqrt{x^2-4}[/tex]
    or [tex]3+\sqrt{x^2-4}= \sqrt{x^2+6x-7}[/tex]
    Square again. The x2 terms cancel and you get a simple equation for x.

    ehild
     
  6. Aug 31, 2012 #5

    SammyS

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    An additional comment:

    Once you do that last step, you will still have a square root. Fortunately at that point, the equation you are left with is fairly simple (uncomplicated).
     
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