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Math Help - Finding X (Quite easy..)

  1. #1
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    Finding X (Quite easy..)

    (3y-1)-1/2(2y+6/8-y-5/3)
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    Quote Originally Posted by Julian.Ignacio View Post
    (3y-1)-1/2(2y+6/8-y-5/3)
    Well, Julian. I give up. Where is x?
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    Quote Originally Posted by masters View Post
    Well, Julian. I give up. Where is x?
    Sorry for misusing some words .
    It's just a monomial calculation, I know the answer, but I forgot how to develop it.
    answer: 73/24y-53/24

    But I wasn't sarcastic, it IS easy. note: /=fraction (I don't know html coding sorry).
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  4. #4
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    Quote Originally Posted by Julian.Ignacio View Post
    (3y-1)-1/2(2y+6/8-y-5/3)
    Quote Originally Posted by Julian.Ignacio View Post
    Sorry for misusing some words .
    It's just a monomial calculation, I know the answer, but I forgot how to develop it.
    answer: 73/24y-53/24

    But I wasn't sarcastic, it IS easy. note: /=fraction (I don't know html coding sorry).
    Your title said to 'find x' and your expression is in 'y'.

    I'm having trouble untangling your expression. Is it one of these:

    (1) 3y-1-\frac{1}{2}\left(\frac{2y+6}{8-y}-\frac{5}{3}\right)

    (2) 3y-1-\frac{1}{2}\left(2y+\frac{6}{8}-y-\frac{5}{3}\right)

    (3) 3y-1-\frac{1}{2}\left(\frac{2y+6}{8}-\frac{y-5}{3}\right)

    Or something else entirely?
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  5. #5
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    Talking

    Quote Originally Posted by Julian.Ignacio View Post
    It's just a monomial calculation, I know the answer.... it IS easy.
    You ask to find x, but you seem to need to simplify an expression in y...?

    You say it's easy, but you show no progress on it...?

    I'm confused.
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  6. #6
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    The third one. Thank you
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  7. #7
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    Quote Originally Posted by stapel View Post
    You ask to find x, but you seem to need to simplify an expression in y...?

    You say it's easy, but you show no progress on it...?

    I'm confused.
    What I meant about easy is that it's basic algebra (11 year old swiss here sorry).
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  8. #8
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    Quote Originally Posted by masters View Post
    Your title said to 'find x' and your expression is in 'y'.

    I'm having trouble untangling your expression. Is it one of these:

    (1) 3y-1-\frac{1}{2}\left(\frac{2y+6}{8-y}-\frac{5}{3}\right)

    (2) 3y-1-\frac{1}{2}\left(2y+\frac{6}{8}-y-\frac{5}{3}\right)

    (3) 3y-1-\frac{1}{2}\left(\frac{2y+6}{8}-\frac{y-5}{3}\right)

    Or something else entirely?
    Quote Originally Posted by Julian.Ignacio View Post
    The third one. Thank you

    (3) 3y-1-\frac{1}{2}\left(\frac{2y+6}{8}-\frac{y-5}{3}\right)


    Simplify what's in parenthesis. The common denominator will be 24.

    3y-1-\frac{1}{2}\left(\frac{3(2y+6)}{24}-\frac{8(y-5)}{24}\right)

    3y-1-\frac{1}{2}\left(\frac{6y+18}{24}-\frac{8y-40}{24}\right)


    3y-1-\frac{1}{2}\left(\frac{6y+18-8y+40}{24}\right)

    3y-1-\frac{1}{2}\left(\frac{-2y+58}{24}\right)


    3y-1+\frac{2y-58}{48}


    Put everything over a denominator of 48.

    \frac{144y-48+2y-58}{48}

    \frac{146y-106}{48}

    \boxed{\frac{73y}{24}-\frac{53}{24}}

    And by the way....Go Federer!
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  9. #9
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    Hahaha pardon for the backled english and thanks for the answer AND Fed xDDD hahahah.
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