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Math Help - Can anybody please show me how to solve this problem step by step?

  1. #1
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    Post Can anybody please show me how to solve this problem step by step?

    [1/(c-2)] - [3c/(c^2 + 2c - 8)] = -4/ (c+4)
    Thanks
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  2. #2
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    Quote Originally Posted by gnarlycarly227 View Post
    [1/(c-2)] - [3c/(c^2 + 2c - 8)] = -4/ (c+4)
    Thanks
    You basically want to factorise c^2 + 2c - 8 to (c+4)(c-2) then cancel out.
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  3. #3
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    1/(c-2) = (c+4)/((c-2)(c+4))

    -4/(c+4) = -4(c-2)/((c-2)(c+4))

    3c/(c^2 + 2c - 8) = 3c/((c-2)(c+4))

    then you have....

    (c+4)-3c+4(c-2)/((c-2)(c+4)) = 0

    2c-4/((c-2)(c+4)) = 0

    2(c-2)/((c-2)(c+4)) = 0

    2/(c+4) = 0
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  4. #4
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    Wait, did you want to solve or to simplify... oops.

    OK, take the line

    (c+4)-3c+4(c-2)/((c-2)(c+4)) = 0

    (c+4)-3c+4(c-2) = 0

    2c-4 = 0

    2c = 4

    c = 2
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