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Math Help - Algebraic Fractions: Is this method correct?

  1. #1
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    Algebraic Fractions: Is this method correct?

    I am having trouble finding the correct method for solving algebraic fractions. I have tried the sequence of operations below and it appears to fit, but would appreciate some help / guidance on a robust process for working through to solve these equations. Thanks

    Solve for x

    x+5/x - 8/2x-3 = 0

    can be rewritten as x+5/x = 8/2x-3

    which cancels out to 5 = 8/2x-3

    which can be rewritten as 8/5 = 2x-3 => 1 3/5 = 2x-3

    and therefore 2x = 1 3/5 + 3

    and x = 1/2(4  3/5)

    ==> x = 2 3/10
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  2. #2
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    It would really help if you wrote the fractions in a non-ambiguous way... did you mean \frac{x+5}{x} -\frac{8}{2x-3} = 0? or x + \frac{5}{x} - \frac{8}{2x} -3 = 0? or something else? It is very hard to understand...

    If you meant the first, then: \frac{x+5}{x} = \frac{8}{2x-3} \Rightarrow (x+5)(2x-3) = 8x \Rightarrow 2x^2 +7x -15 = 8x \Rightarrow 2x^2 -x -15 = 0

    Now solve for x and that is your solution (use the quadratics formula x_{1,2} = \frac{-b \pm \sqrt{b^2-4ac} }{2a})
    Last edited by Defunkt; October 24th 2009 at 08:28 AM.
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  3. #3
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    Thank you for replying, (you read the equation correctly) but your response is a little unclear beyond the middle section.


    Cross multiplying to obtain (x+5)(2x-3) = 8x makes perfect sense, as does expanding the brackets on the left side. However I can't follow how 8x translates to 16x - 24

    I appears as if the 8x value is discarded, and 8 is re multiplied against 2x-3

    Would this be what it is supposed to look like:

    <br />
\frac{(x+5)(2x-3)}{x} = \frac{8(2x-3)}{2x-3}

    and could you clarify this?

    Thanks
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  4. #4
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    <br />
\frac{x+5}{x} =\frac{8}{2x-3}<br />

    Cross multiplying gives you:

    <br />
(x+5)(2x-3) = 8x<br />

    Expanding the LHS and collecting like terms

    <br />
2x^2-3x+10x-15= 8x \Rightarrow 2x^2-x-15=0<br />

    Use the quadratic formula to find the solutions.

    Which are  x= -\frac{5}{2} or x=3
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  5. #5
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    Quote Originally Posted by vbiter View Post
    I am having trouble finding the correct method for solving algebraic fractions. I have tried the sequence of operations below and it appears to fit, but would appreciate some help / guidance on a robust process for working through to solve these equations. Thanks

    Solve for x

    x+5/x - 8/2x-3 = 0

    can be rewritten as x+5/x = 8/2x-3

    which cancels out to 5 = 8/2x-3
    No, it doesn't. If you mean (x+5)/x, that is equal to 1+ 5/x, not 5.

    Assuming that you mean (x+5)/x= 8/(2x-3), I would recommend that you get rid of the fractions by multiplying both sides by the denominators x(2x-3)

    Multiplying the left side by x(2x-3) cancels the "x"s and gives (x+5)(2x-3). Multiplying the right side by x(2x-3) cancels the "2x-3"s and gives 8(x). So your equation becomes (x+5)(2x-3)= 8x. That is a quadratic equation for x.

    which can be rewritten as 8/5 = 2x-3 => 1 3/5 = 2x-3

    and therefore 2x = 1 3/5 + 3

    and x = 1/2(4  3/5)

    ==> x = 2 3/10
    Last edited by mr fantastic; October 24th 2009 at 05:11 PM. Reason: Added close quote tag
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  6. #6
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    Quote Originally Posted by vbiter View Post
    Thank you for replying, (you read the equation correctly) but your response is a little unclear beyond the middle section.


    Cross multiplying to obtain (x+5)(2x-3) = 8x makes perfect sense, as does expanding the brackets on the left side. However I can't follow how 8x translates to 16x - 24

    I appears as if the 8x value is discarded, and 8 is re multiplied against 2x-3

    Would this be what it is supposed to look like:

    <br />
\frac{(x+5)(2x-3)}{x} = \frac{8(2x-3)}{2x-3}

    and could you clarify this?

    Thanks
    Oops, don't know what happened there... fixed the original post now.
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