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Solving an equation

  1. Jul 28, 2008 #1

    should give x=[tex]\frac{1+c}{2-a}[/tex]

    I've been trying to solve it now for like two hours. Some hints?

    thank you
  2. jcsd
  3. Jul 28, 2008 #2


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    Staff: Mentor

    Show what you did, it is hard to tell where is your problem. In general equation is not difficult - multiply, rearrange and you should get there.
  4. Jul 28, 2008 #3

    How do I get rid of the a² especially?
  5. Jul 28, 2008 #4


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    So x- 1/4 a2x= (1- 1/4 a)x= 1/2+ 1/2 c+ 1/4 a+ 1/4 ac
    Now what do you do?
  6. Jul 28, 2008 #5


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    Staff: Mentor

    More like

    [tex]x-\frac 1 4 a^2x = (1- \frac 1 4 a^2)x[/tex]
  7. Jul 28, 2008 #6
    Dividing by a?

    (1/a-a/4)x=1/2a + c/2a + 1/4 + c/4

    but now?
  8. Jul 28, 2008 #7


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    NO! Divide by (1- 1/4 a2)!
  9. Jul 28, 2008 #8
    Why should I do that?

    Nothing cancels then on the RHS?
  10. Jul 28, 2008 #9


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    Staff: Mentor

    Divide, you will be surprised how many things cancel out :smile:


    [tex]a^2 - b^2 = (a-b)(a+b)[/tex]
  11. Jul 28, 2008 #10
    Got it! thanks, guys!
  12. Jul 28, 2008 #11
    I am getting x = (1+c) / (1/2-a)

    Is everyone else getting the "2-a" as the denominator as it was stated in the solution posted?
  13. Jul 28, 2008 #12
    Never-mind. I got "2-a" as my denominator now :) .

    I didn't think to separate the "1" into "(2(1/2))", so that the "1/2's" could cancel out. Although, if I had not done that, and I just divided the "1/2" in the numerator, by the "1" in the denominator, to get "1/2-a", why exactly would it have been wrong?

    This was the set up that I had....

    x= (1/2 + 1/4a) (1+c) / (1-1/4a^2) ....

    Then I rearranged the denominator so that values could cancel out....

    x= (1/2 + 1/4a) (1+c) / (2(1/2)) - a(1/4a) ....where the "1/2" and the "1/4a" cancel out....

    but again...what if I had changed the denominator to "1 - a(1/4a)" and got "1/2-a"?

    Thanks to anyone else in advance...I hope I was clear enough with my question.

  14. Jul 29, 2008 #13


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    Staff: Mentor

    Where did you get 1/2 from?

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

    It is much simpler IMHO to start by multiplying both sides of the equation by 4 at some early stage - then you end without fractions.
  15. Jul 29, 2008 #14
    Sorry to waste your time Borek. I noticed another error I made in my earlier calculations which screwed the whole problem up (so everyone, ignore my previous posts! haha). Since then I have fixed it and have the correct solution. Thank you anyway for your help!
  16. Jul 29, 2008 #15
    Why all the superfluous algebraic manipulation and rearranging? Just consider the root equation satisfied here:

    x = (1+c+ax)/2

    Solving for x gives the desired conclusion immediately.
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