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Math Help - Distributing with fractions

  1. #1
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    Distributing with fractions

    I don't understand the rules for distributing with fractions in problems like these:

    2/3 (c + 7)

    -5/8 (k + 8)

    24 (x/3 - 1/8)

    I'd greatly appreciate help on these specific problems, and I would also appreciate if someone could explain the rules to me.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by endlesst0m View Post
    I don't understand the rules for distributing with fractions in problems like these:

    2/3 (c + 7)
    the distributive law states:

    a(b + c) = ab + ac

    in other words, the term that is multiplying a sum is distributed and multiply each term in the sum.

    so here, \frac 23 (c + 7) = \frac 23 \cdot c + \frac 23 \cdot 7 = \frac {2c}3 + \frac {14}3

    you can combine these, but if tht is how you wanted the answer, the distributive law could be applied differently.

    \frac 23 (c + 7) = \frac {2(c + 7)}3 = \frac {2c + 14}3

    now try the others. if it helps, you can think of 24 as \frac {24}1
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  3. #3
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    So basically, when you distribute a fraction into a variable, the numerator in the fraction combines with the variable, and the denominator stays the same.

    And when you distribute a fraction into a number, you multiply that number times the numerator, and the denominator stays the same.

    Right?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by endlesst0m View Post
    So basically, when you distribute a fraction into a variable, the numerator in the fraction combines with the variable, and the denominator stays the same.

    And when you distribute a fraction into a number, you multiply that number times the numerator, and the denominator stays the same.

    Right?
    no, you multiply each term. look at the rule. it is a coincidence that the denominator stays the same in this case.

    recall how to multiply fractions: \frac ab \cdot \frac cd = \frac {ac}{bd}

    the denominator in this case stays the same, because if you write c + 7 as a fraction, it would be \frac {c + 7}1. so when you multiply the denominators, it keeps the denominator of the fraction outside, because you are multiplying it by 1
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