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Math Help - I can't do this algebra poblem

  1. #1
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    I can't do this algebra poblem

    I can't do this algebra problem can anyone help please

    (a) Write down an expression of the form

    __k__ __K^2__
    ax+b + cx+d

    where a and c are non-zero, and the denominators are distinct

    (b) Add up the terms of the previous part of the equation to obtain a single fraction.

    I'm spending way too long at this question and i know i'm doing it all wrong
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  2. #2
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    hello
    would u please show us your work,so we can see where you went wrong
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  3. #3
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    ok it might take me a while to type it out bare with me
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  4. #4
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    __k__ + __k^2__
    ax+b cx+b

    i put a=2 and c=3

    __k__ + __k^2__
    2x+b 3x+d

    _k(3x+d)_ + _k^2(2x+b)_
    (2x+b)(3x+d) (2x+b)(3x+d)

    __k(3x+d)+k^2(2x+b)__
    (2x+b)(3x+d)

    i know i've prob gone badly wrong
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  5. #5
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    u didn't go wrong,
    \frac{k}{2x+b}+\frac{k^2}{3x+d}=\frac{k(3x+d)}{(2x  +b)(3x+d)}+ \frac{k^2(2x+b)}{(3x+d)(2x+b)} =\frac{k(3x+d)+k^2(2x+b)}{(2x+b)(3x+d)}
    that is right.
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  6. #6
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    oh right so what do i do next to finish the question??
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  7. #7
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    well,i didn't get the question quite good,i hope someone else would help you solve it.
    Last edited by Raoh; November 29th 2009 at 07:03 AM. Reason: didn't get it
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  8. #8
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    is it right to have 3kx+kd+2k^2x+k^2b as my top line?
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  9. #9
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    Quote Originally Posted by carly1990 View Post
    is it right to have 3kx+kd+2k^2x+k^2b as my top line?
    Yes that is right to have  3kx+kd+2xk^2+bk^2 as a top line.
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  10. #10
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    iv'e got it down too

    3kx+xd+2xk^2+bk^2
    --------------------
    6x^2+xb+2xd+db

    i haven't a clue what to do now
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  11. #11
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    3kx+xd+2xk^2+bk^2
    --------------------
    6x^2+3xb+2xd+db

    sorry thats what i got haven't a clue what to do now
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  12. #12
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    Quote Originally Posted by carly1990 View Post
    3kx+xd+2xk^2+bk^2
    --------------------
    6x^2+3xb+2xd+db

    sorry thats what i got haven't a clue what to do now
    what do u want to do exactly ?
    ('cause i don't have a clue either )
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  13. #13
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    Quote Originally Posted by Raoh View Post
    u didn't go wrong,
    \frac{k}{2x+b}+\frac{k^2}{3x+d}=\frac{k(3x+d)}{(2x  +b)(3x+d)}+ \frac{k^2(2x+b)}{(3x+d)(2x+b)} =\frac{k(3x+d)+k^2(2x+b)}{(2x+b)(3x+d)}
    that is right.
    we already have a single fraction here, \left [\frac{k(3x+d)+k^2(2x+b)}{(2x+b)(3x+d)}  \right ]
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  14. #14
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    so the answer to (a) and (b) are the same??

    (a) Write down an expression of the form

    __k__ __K^2__
    ax+b + cx+d

    where a and c are non-zero, and the denominators are distinct

    (b) Add up the terms of the previous part of the equation to obtain a single fraction.
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  15. #15
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    Quote Originally Posted by carly1990 View Post
    so the answer to (a) and (b) are the same??

    (a) Write down an expression of the form

    __k__ __K^2__
    ax+b + cx+d

    where a and c are non-zero, and the denominators are distinct

    (b) Add up the terms of the previous part of the equation to obtain a single fraction.
    for (a) i think you must give some values to a,b,c and d ( a\neq 0 , c\neq 0 ) such that ax+b\neq cx+d and k is any real number(better give it a value too).for (b),see my second post.
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