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Equation solving/solving for x

  1. Oct 13, 2015 #1
    1. The problem statement, all variables and given/known data
    1 / 2x + 1 − 2x / x − 3 = x + 1 / 2x ^ 2 − 5x − 3

    solve for x. check solution in original equation

    3. The attempt at a solution

    I've tried working this out a couple different ways. All I can change it to is 4x^2 - x - 3 / 2x^2 - 5x - 3 = x+1/2x^2-5x-3.

    I really don't know where to go or what to do.
    I even put the answer into maple and tried to work it backwards.
    Super stuck. thanks for the help in advance.
    Last edited by a moderator: Oct 13, 2015
  2. jcsd
  3. Oct 13, 2015 #2


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    Homework Helper
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    Gold Member

    Maybe it is not written correctly.
  4. Oct 13, 2015 #3


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    I guess you omitted a few couples of parentheses .
    What you wrote is equivalent to 0.5 x -2 -3 =x+0.5 x^2-5x-3 - are you sure you wanted to write that?
    Last edited by a moderator: Oct 13, 2015
  5. Oct 13, 2015 #4


    Staff: Mentor

    As ehild already said, you need quite a few more parentheses here.

    What you wrote is properly interpreted to mean this:
    ##\frac 1 2 x + 1 - \frac{2x}{x} - 3 = x + \frac 1 2 x^2 - 5x - 3##
    Again, you need some parentheses, plus show your intervening steps in getting to the above.

    Regarding parentheses, when you write fractions on a single line, if the numerator or denominator has more than a single term, put parentheses around it.
    For example: ##\frac{x^2 + 3x + 2}{x + 1}## should be written as (x^2 + 3x + 2)/(x + 1), if written on a single line.
  6. Oct 13, 2015 #5


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    Using parentheses, (4x^2- x- 3)/(2x^2- 5x- 3)= (x+ 1)/(2x^2- 5x- 3)
    Note that the denominator is 0 for x= 3 and x= -1/2. As long as the denominator is NOT 0, we can multiply on both sides by it to get
    4x^2- x- 3= x+ 1 which is the same as the quadratic equation 4x^2- 2x- 4= 0
  7. Oct 13, 2015 #6
    From where you are you could put all on the same side since you have a common denominator you can put it away and remains a quadratic equation
  8. Oct 24, 2015 #7
    Thanks for the help everyone!
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