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Math Help - i can't figure out how the teacher got 7x..please help

  1. #1
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    Unhappy i can't figure out how the teacher got 7x..please help

    ok so i got this question and i know how to slove it and everything but im lost as of where the teacher got this one number.. so here it is


    (x+5)/(x-2) = 5/(x+2) + 28/(x^2-4)


    so from that i got ..

    x^2+10 = 5x-10+28

    but.. the teacher got

    x^2 + 7x +10 = 5x-10+28

    where did he get 7x from???
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  2. #2
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    Quote Originally Posted by Candy101 View Post
    ok so i got this question and i know how to slove it and everything but im lost as of where the teacher got this one number.. so here it is


    (x+5)/(x-2) = 5/(x+2) + 28/(x^2-4)

    x^2 + 7x +10 = 5x-10+28

    where did he get 7x from???
    Hmm, not sure myself. If i do the calculation for the left part

    1. (x+5) multiplied by (x-2) since in order to get rid of the denumerator
    2. gettin rid of brackets
    3. x^2 -2X + 5X -10

    I get +3x .. lol
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  3. #3
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    Quote Originally Posted by Candy101 View Post
    ok so i got this question and i know how to slove it and everything but im lost as of where the teacher got this one number.. so here it is


    (x+5)/(x-2) = 5/(x+2) + 28/(x^2-4)


    so from that i got ..

    x^2+10 = 5x-10+28

    but.. the teacher got

    x^2 + 7x +10 = 5x-10+28

    where did he get 7x from???
    Note that due to the difference of two squares: (x^2-4)=(x-2)(x+2). Also note that the best method is to get the same denominator for different terms, in this case it makes sense to use x^2-4 as it is the LCM of x-2 and x+2

    Multiply the first term by \frac{x+2}{x+2}

    Multiply the second term by \frac{x-2}{x-2}

    This should now give each term a denominator of x^2-4 which cancels to

    (x+5)(x+2) = 5(x-2)+28

    x^2+7x+10 = 5x+18<br />

    For this to work x \neq \pm 2

    edit: see spoiler for detailed instructions

    Spoiler:
    \frac{x+5}{x-2} = \frac{5}{x+2} + \frac{28}{x^2-4} = \frac{x+5}{x-2} = \frac{5}{x+2} + \frac{28}{(x-2)(x+2)}

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

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

    Combine all these now

    \frac{(x+5)(x+2)}{(x-2)(x+2)} = \frac{5(x-2)}{(x-2)(x+2)} + \frac{28}{x^2-4}

    \frac{(x+5)(x+2)}{(x-2)(x+2)} = \frac{5(x-2)}{(x-2)(x+2)} + \frac{28}{(x-2)(x+2)}

    (x-2)(x+2) will cancel

    (x+5)(x+2) = 5(x-2) + 28

    Expand using FOIL

    x^2+5x+2x+10 = 5x-10+28

    Simplify

    x^2+7x+10=5x+18
    Last edited by e^(i*pi); December 6th 2009 at 09:57 AM.
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  4. #4
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    Quote Originally Posted by Candy101 View Post
    ok so i got this question and i know how to slove it and everything but im lost as of where the teacher got this one number.. so here it is


    (x+5)/(x-2) = 5/(x+2) + 28/(x^2-4)


    so from that i got ..

    x^2+10 = 5x-10+28

    but.. the teacher got

    x^2 + 7x +10 = 5x-10+28

    where did he get 7x from???
    Hi Candy101,

    \frac{x+5}{x-2}=\frac{5}{x+2}+\frac{28}{x^2-4}

    First, multiply through by the common denominator, which is (x - 2)(x + 2).

    (x+5)(x+2)=5(x-2)+28

    Expand and set everything equal to zero. This is where you went wrong. You multiplied the first two binomials incorrectly.

    x^2+7x+10=5x-10+28

    x^2+7x+10-5x+10-28=0

    x^2+2x-8=0

    Factor.

    (x+4)(x-2)=0

    x=-4

    x \ne 2 because it would cause the denominator to be zero in the first fraction.
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  5. #5
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    hey thanks guys for the help!!!!!!!!!!!!!!!!!!!!!!!!!!
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