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Math Help - Variable Factions help

  1. #1
    Member cmf0106's Avatar
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    Variable Factions help

    \frac{2}{15x} + \frac{7}{18}

    Can someone please go through the details on how to solve this? The correct answer is  \frac{12+35x}{90x}

    Edit: Title should read variable Fractions obviously my mistake.
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    Quote Originally Posted by cmf0106 View Post
    \frac{2}{15x} + \frac{7}{18}

    Can someone please go through the details on how to solve this? The correct answer is  \frac{12+35x}{90x}

    Edit: Title should read variable Fractions obviously my mistake.
    15x=3\cdot 5 \cdot x
    18=2 \cdot 3^2

    LCD=90x

    \frac{90x(\frac{2}{15x})}{90x}+\frac{90x(\frac{7}{  18})}{90x}

    \frac{12}{90x}+\frac{35x}{90x}=\frac{12+35x}{90x}
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  3. #3
    Member cmf0106's Avatar
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    I think that clarifies that question a few more on the same subject though.

    1.in this example for canceling variables we have
    \frac{2+x}{3} = \frac{2}{x} + \frac{x}{x} = \frac{2}{x} +1

    Note that in this example the x in the denominator is applied to the 2 and x in the numerator. Shown here  \frac{2+x}{x}, the x in the denominator goes to both numerators thus \frac{2}{x} + \frac{x}{x}

    2.However in this example
    \frac{(2 + 3x)(x-1)}{2 + 3x} For the solution the author does not apply the whole denominator to both of the numerators and this is confusing me greatly. Working this problem out we have \frac{(2 + 3x)(x-1)}{2 + 3x}=\frac{2+3x}{2+3x} * \frac{x-1}{1}= x -1

    why is the 2 + 3x not applied to both parts of this numerator resulting in \frac{2+3x}{2+3x} * \frac{x-1}{2+3x}?

    thanks in advance
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    A riddle wrapped in an enigma
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    Quote Originally Posted by cmf0106 View Post
    I think that clarifies that question a few more on the same subject though.

    1.in this example for canceling variables we have
    \frac{2+x}{\color{red}3} = \frac{2}{x} + \frac{x}{x} = \frac{2}{x} +1

    Note that in this example the x in the denominator is applied to the 2 and x in the numerator. Shown here  \frac{2+x}{x}, the x in the denominator goes to both numerators thus \frac{2}{x} + \frac{x}{x}

    I assume you intended to put an x in the denominator instead of a 3.

    2.However in this example
    \frac{(2 + 3x)(x-1)}{2 + 3x} For the solution the author does not apply the whole denominator to both of the numerators and this is confusing me greatly. Working this problem out we have \frac{(2 + 3x)(x-1)}{2 + 3x}=\frac{2+3x}{2+3x} * \frac{x-1}{1}= x -1

    why is the 2 + 3x not applied to both parts of this numerator resulting in \frac{2+3x}{2+3x} * \frac{x-1}{2+3x}?

    thanks in advance
    The first expression is made up of addends (terms that are added together). You must have a common denominator to combine them.

    Your second expression is a product. You do not need a common denominator. You simply multiply numerators and denominators and factor out any common elements in the product or along the way.

    \frac{(2 + 3x)(x-1)}{2 + 3x}=\frac{2+3x}{2+3x} * \frac{x-1}{1}=\not[\frac{2+3x}{2+3x}] * \frac{x-1}{1}= \frac{1}{1} * \frac{x-1}{1}=x -1

    \frac{2+3x}{2+3x} reduces to \frac{1}{1}
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  5. #5
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    Quote Originally Posted by masters View Post
    The first expression is made up of addends (terms that are added together). You must have a common denominator to combine them.

    Your second expression is a product. You do not need a common denominator. You simply multiply numerators and denominators and factor out any common elements in the product or along the way.

    \frac{(2 + 3x)(x-1)}{2 + 3x}=\frac{2+3x}{2+3x} * \frac{x-1}{1}=\not[\frac{2+3x}{2+3x}] * \frac{x-1}{1}= \frac{1}{1} * \frac{x-1}{1}=x -1

    \frac{2+3x}{2+3x} reduces to \frac{1}{1}
    Look at a numerical example:

    \frac{(2)(3)}{6} \neq \frac{2}{6} * \frac{3}{6} \neq \frac{6}{36} \neq \frac{1}{6}

    Instead, it's this:

    \frac{(2)(3)}{6}=\frac{2}{6} * \frac{3}{1} = \frac{6}{6} = 1

    Generally,

    \frac{ab}{c}=\frac{a}{c}*\frac{b}{1}

    If we did it as you suggest, we would have

    \frac{ab}{c}=\frac{a}{c} * \frac{b}{c} = \frac{ab}{c^2} \neq \frac{ab}{c}
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