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Math Help - Rational Problems

  1. #1
    Ari
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    Rational Problems

    Okay so i dont understand how to do this problem...

    x+6 + x+3
    x2+x 12 x2-16

    (x2=x squared)

    Im supposed to simplyfy and state the restrictions on the variables
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Ari View Post
    Okay so i dont understand how to do this problem...

    x+6 + x+3
    x2+x-12 x2-16

    (x2=x squared)
    you are supposed to be simplifying i suppose? please state your questions clearly and specifically.

    factor the denominators and then combine the fractions.

    do you know how to factor?
    do you know how to combine fractions?
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  3. #3
    Ari
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    Im supposed to simplify it and then state the restrictions
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Ari View Post
    Im supposed to simplify it and then state the restrictions
    ok, so try what i said
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  5. #5
    Ari
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    Ok So how would i do


    5m(squared) +6m-8 * |20m(squared) +16m
    5m(squared) -16m-16 |4-m (squared)

    |mean the other fraction
    Last edited by Ari; September 30th 2007 at 12:20 PM.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Ari View Post
    Ok So how would i do

    2 2
    5m +6m-8 * 20m +16m
    2 2
    5m -16m-16 4-m
    2
    And btw those 2's on top mean squared so 5m mean 5m to the power of 2
    i don't understand what you are saying. if you can't use LaTex, please write fractions as follows:

    (numerator)/(denominator)

    include the brackets

    what did you get for the first question
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  7. #7
    Ari
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    I got 2x(squared)+2x-33/(x+4)(x-4)(x-3) for the first one
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Ari View Post
    I got 2x(squared)+2x-33/(x+4)(x-4)(x-3) for the first one
    please rewrite the original question so i can check it. there is a sign missing from the 12 in the original post
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  9. #9
    Ari
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    (x+6)/(x^2+x-12) X (x+3)/(x^2-16)
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    Quote Originally Posted by Ari View Post
    (x+6)/(x^2+x-12) X (x+3)/(x^2-16)
    if it was x^2 {\color{red}-}x - 12 in the denominator of the first fraction, we would be in business. but since it is + x, nothing cancels. we get: \frac {x + 6}{x^2 + x - 12} \cdot \frac {x + 3}{x^2 - 16} = \frac {(x + 6)(x + 3)}{(x + 4)^2 (x - 4)(x + 3)}

    the restrictions are, we cannot choose the x's that make the denominator zero. what x's are those?
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  11. #11
    Ari
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    Quote Originally Posted by Jhevon View Post
    if it was x^2 {\color{red}-}x - 12 in the denominator of the first fraction, we would be in business. but since it is + x, nothing cancels. we get: \frac {x + 6}{x^2 + x - 12} \cdot \frac {x + 3}{x^2 - 16} = \frac {(x + 6)(x + 3)}{(x + 4)^2 (x - 4)(x + 3)}

    the restrictions are, we cannot choose the x's that make the denominator zero. what x's are those?

    could you show the steps for that? I thought we are supposed to multyply both bottomt and top when doing the LCD
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Ari View Post
    could you show the steps for that? I thought we are supposed to multyply both bottomt and top when doing the LCD
    you had a multiplication sign, we do not fin any LCD's here. we just multiply the tops to get the new top, and multiply the bottoms to get the new bottom
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  13. #13
    Ari
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    Uggh sorry i messed up its supposed to be a + sign in the middle of the fractions

    (x+6)/(x^2+x-12) + (x+3)/(x^2-16)
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    Quote Originally Posted by Ari View Post
    Uggh sorry i messed up its supposed to be a + sign in the middle of the fractions

    (x+6)/(x^2+x-12) + (x+3)/(x^2-16)
    ok.


    \frac {x + 6}{x^2 + x - 12} + \frac {x + 3}{x^2 - 16} = \frac {x + 6}{(x + 4)(x - 3)} + \frac {x + 3}{(x + 4)(x - 4)}

    LCD here is: (x + 4)(x - 4)(x - 3)

    so we have: \frac {x + 6}{(x + 4)(x - 3)} + \frac {x + 3}{(x + 4)(x - 4)} = \frac {(x + 6)(x - 4) + (x + 3)(x - 3)}{(x + 4)(x - 4)(x - 3)}

    = \frac {x^2 + 2x - 24 + x^2 - 9}{(x + 4)(x - 4)(x - 3)}

    = \frac {2x^2 + 2x - 33}{(x + 4)(x - 4)(x - 3)}

    that cannot be factored any further.

    our restrictions are: x \ne \pm 4 and x \ne 3 since those x's make the denominator zero
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  15. #15
    Ari
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    Quote Originally Posted by Jhevon View Post
    ok.


    \frac {x + 6}{x^2 + x - 12} + \frac {x + 3}{x^2 - 16} = \frac {x + 6}{(x + 4)(x - 3)} + \frac {x + 3}{(x + 4)(x - 4)}

    LCD here is: (x + 4)(x - 4)(x - 3)

    so we have: \frac {x + 6}{(x + 4)(x - 3)} + \frac {x + 3}{(x + 4)(x - 4)} = \frac {(x + 6)(x - 4) + (x + 3)(x - 3)}{(x + 4)(x - 4)(x - 3)}

    = \frac {x^2 + 2x - 24 + x^2 - 9}{(x + 4)(x - 4)(x - 3)}

    = \frac {x^2 + 2x - 33}{(x + 4)(x - 4)(x - 3)}

    that cannot be factored any further.

    our restrictions are: x \ne \pm 4 and x \ne 3 since those x's make the denominator zero

    YOUR AWSOME!! but shouldnt it be 2x^2??
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