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Math Help - algebraic fraction 2

  1. #1
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    algebraic fraction 2

    Can any one help with two problems I have


    x/xy+x^2 + y/x^2+xy =x/y(x+y) +y/x(x+y)=x^2/xy(x+y)+y^2/xy(x+y)=x^2+y^2/xy(x+y)

    with this ome I can work out why xy(x+y) s the common factor and why the x and th Y become squared.


    1/x^2+x - 1/x+1 = 1/x(x+1) -x/x(x+1)

    with this one why does the x appear as a numerator


    hope this all makes sence.

    many thanks

    Dave
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  2. #2
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    Re: algebraic fraction 2

    Quote Originally Posted by davellew69 View Post
    Can any one help with two problems I have


    x/xy+x^2 + y/x^2+xy =x/y(x+y) +y/x(x+y)=x^2/xy(x+y)+y^2/xy(x+y)=x^2+y^2/xy(x+y)

    with this ome I can work out why xy(x+y) s the common factor and why the x and th Y become squared.


    1/x^2+x - 1/x+1 = 1/x(x+1) -x/x(x+1)

    with this one why does the x appear as a numerator


    hope this all makes sence.

    many thanks

    Dave
    because ...

    \frac{1}{x+1} \cdot \frac{x}{x} = \frac{x}{x(x+1)}



    next time, use grouping symbols to collect terms in the numerator and denominator of your fractions

    for example, x/xy+x^2 could be misinterpreted as \frac{x}{xy} + x^2 , so write it as x/(xy+x^2)
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  3. #3
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    Re: algebraic fraction 2

    thanks for the reply but I `m lost on what your reply was trying to say

    dave
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  4. #4
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    Re: algebraic fraction 2

    Quote Originally Posted by davellew69 View Post
    thanks for the reply but I `m lost on what your reply was trying to say

    dave
    you originally asked ...

    1/x^2+x - 1/x+1 = 1/x(x+1) -x/x(x+1)
    with this one why does the x appear as a numerator
    I answered your question ... note that the common denominator of the two fractions is x(x+1).
    Thanks from mash
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  5. #5
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    Re: algebraic fraction 2

    The "1/x(x+1) - x/x(x+1)" in previous post needs further bracketing:
    1 / [x(x+1)] - x / [x(x+1)]

    Your starting expression should be shown this way:
    1 / (x^2 + x) - 1 / (x + 1) ; then:

    = 1 / [x(x + 1)] - 1 / (x + 1)

    = 1 / [x(x+1)] - x / [x(x+1)]

    = (1 - x) / [x(x + 1)]
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  6. #6
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    Re: algebraic fraction 2

    Quote Originally Posted by davellew69 View Post
    x/xy+x^2 + y/x^2+xy =x/y(x+y) +y/x(x+y)=x^2/xy(x+y)+y^2/xy(x+y)=x^2+y^2/xy(x+y)
    Dave, that's quite messy; if you don't show proper bracketing, then I don't think we should
    lose our time trying to decipher what you mean. Your original expression "x/xy+x^2 + y/x^2+xy"
    MUST be shown this way:
    x / (xy + x^2) + y / (x^2+xy) ; and

    x / (xy + x^2) = x / [x(x + y)] , not x / [y(x + y)] as you have...

    For your benefit:
    20 / (2 + 3) = 20 / 5 = 4
    20 / 2 + 3 = 10 + 3 = 13
    See the importance of proper bracketing?
    Thanks from mash and anonimnystefy
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