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Algebra, desperate

  1. Feb 7, 2008 #1
    1. The problem statement, all variables and given/known data
    6+X = 3+(X/2)

    2. Relevant equations
    Quadratic formula?

    3. The attempt at a solution
    2(6+X) = 3+X

    12+2X = 3+X

    12+2X = X(3+X)

    12+2X = 3X-X^2



    when putting it in the quadratic formula, I don't get the right answers.
    the answer should be X1= 2 X2= -6
    what have I done wrong? :(
    Last edited: Feb 7, 2008
  2. jcsd
  3. Feb 7, 2008 #2
    You're messing up your operations. Go back to the first line in your attempt. The RHS is 3 + x/2 NOT (3 + x)/2

    Try again
  4. Feb 7, 2008 #3


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    Starting with (X+6)/X=3+X/2, if you multiply both sides by 2 you get (2X+12)/X=6+X. Note the 6 instead of a 3. 3+X/2 is not the same as (3+X)/2.
  5. Feb 7, 2008 #4
    I haven't followed the whole thing through. But it looks like your first move is to multiply both sides by 2 and you got the wrong answer on the right side of the equation.

    You wouldn't get 3+X, you'd get 2(3+X/2).

    For the first step, you need to make a common denominator (2) on the right side before you multiply both sides by two. Does that make sense?

    Or, do it Dick's way.
  6. Feb 7, 2008 #5
    Thank you very much!!! I got the right answer now. Am I supposed to grade the help I got? If so, how?

    I'll probably have some more questions tonight... can I just post it in here or should I create a new thread?
  7. Feb 7, 2008 #6


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    Grades? Grade US? We don't need no steenking grades! :) If it's a new question start a new thread, old long threads tend to get overlooked.
  8. Feb 7, 2008 #7
    Oh, thought you were supposed to grade it as in some swedish forums :D.
    Well, thank you again!
  9. Feb 7, 2008 #8
    If you want, give out ATTABOYs. When posters get 20 ATTABOYs, it cancels out one MY BAD!
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