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Math Help - Complex Fractions with variable in denominator

  1. #1
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    Complex Fractions with variable in denominator

    2 / 5 / x + 4

    Having trouble solving this problem and have no clue where to start. I'm trying to eliminate any fractions in the denominator.

    Couln't figure out how to format this right so I just have it typed out. It's 2 divided by 5 over x + 4
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  2. #2
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    e^(i*pi)'s Avatar
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    Use \dfrac for a fraction: \dfrac{2}{5x+4}

    However, I see no imaginary part and hence no complex number here. That form is the simplest it's going to get since you can't eliminate x from the denominator.

    What are you hoping to achieve?

    EDIT:

    Assuming you mean \dfrac{2}{\left(\frac{5}{x+4}\right)} then recall that dividing by a fraction is the same as flipping and multiplying it. Thus \dfrac{2}{\left(\frac{5}{x+4}\right)} = 2 \times \dfrac{x+4}{5} which has no x in the denominator.
    Last edited by e^(i*pi); February 15th 2011 at 02:53 PM. Reason: OP edited the OP
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by street1030 View Post
    2 / \fraction{5}{x + 4}

    Having trouble solving this problem and have no clue where to start. I'm trying to eliminate any fractions in the denominator.
    \displaystyle \frac{2}{ \frac{5}{x + 4}} Is this it?

    Like any other complex fraction you need to clear the fraction out of the denominator by multiplying the top and bottom of the fraction by the factor that clears the denominator.

    \displaystyle = \left ( \frac{2}{ \frac{5}{x + 4}} \right ) \cdot \frac{x + 4}{x + 4}

    \displaystyle = \frac{2(x + 4)}{ \frac{5}{x + 4} \cdot (x + 4)}

    You finish up from here.

    -Dan
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  4. #4
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    Quote Originally Posted by street1030 View Post
    2 / 5 / x + 4
    Couln't figure out how to format this right so I just have it typed out. It's 2 divided by 5 over x + 4
    \dfrac{\frac{2}{5}}{x+4}=\dfrac{2}{5x+20}
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  5. #5
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    Sorry for the confusion guys. Forgot to preview post before posting. It's supposed to be \dfrac{2}{5} all over x + 4
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    e^(i*pi)'s Avatar
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    That would be Plato's answer then and if you're wondering he/she has "dropped" the 5 into the denominator and then distributed it
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    Having trouble understanding how we went from \displaystyle = \frac{2(x + 4)}{ \frac{5}{x + 4} \cdot (x + 4)}

    to Platos

    \dfrac{2}{5x+20}

    Did you cross multiply to cancel out the x + 4 in the denominator and the numerator. Then just distribute the 5 to the remaining x + 4?
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  8. #8
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    Quote Originally Posted by street1030 View Post
    Did you cross multiply to cancel out the x + 4 in the denominator and the numerator. Then just distribute the 5 to the remaining x + 4?
    \dfrac{\frac{2}{5}}{x+4}=\dfrac{5\left(\frac{2}{5}  \right)}{5(x+4)}=\dfrac{2}{5x+20}
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  9. #9
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    Alright thanks guys! Going to go test this out on some more problems.
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  10. #10
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by street1030 View Post
    Having trouble understanding how we went from \displaystyle = \frac{2(x + 4)}{ \frac{5}{x + 4} \cdot (x + 4)}

    to Platos

    \dfrac{2}{5x+20}

    Did you cross multiply to cancel out the x + 4 in the denominator and the numerator. Then just distribute the 5 to the remaining x + 4?
    They are not the same. I was using a different fraction than the one that Plato simplified for you.

    -Dan
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  11. #11
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    O yeah. I need to do the one you're doing topsquark. So the way you did it. you get? 2x + 8 / 5?
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  12. #12
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by street1030 View Post
    O yeah. I need to do the one you're doing topsquark. So the way you did it. you get? 2x + 8 / 5?
    (2x + 8)/5. Yes. Please note my use of the parenthesis. What you wrote is 2x + (8/5).

    -Dan
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  13. #13
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    Thanks a bunch
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