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Thread: Word Problem: Computing age comparing quantities

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    Word Problem: Computing age comparing quantities

    Problem: Monica earns three times per hour as John, John earns $2 more per hour than Alicia. Together they earn $43 per hour. How much is each one's hourly wage?

    I attempted to compare each quantity in comparison to Monica

    Monica = $\displaystyle x$
    John = Monica earns three times per hour as John, therefore John earns 1/3 in comparison to Monica $\displaystyle = \frac{1}{3x}$
    Alicia = receives $2 less than John = $\displaystyle = \frac{1}{3x}-2$

    The equation then becomes

    $\displaystyle x + (\frac{1}{3x})+ (\frac{1}{3x} -2) = 43$

    Clear out the fractions by multiplying every item by the LCD of 3, so on so forth but I do not get the correct answer.

    The correct answer is $\displaystyle x = 7$, did I make an error along the way or is all of the work completely wrong?

    Thanks for your help
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    Re: Word Problem: Computing age comparing quantities

    Quote Originally Posted by allyourbass2212 View Post
    Problem: Monica earns three times per hour as John, John earns $2 more per hour than Alicia. Together they earn $43 per hour. How much is each one's hourly wage?
    Make it easy on yourself. Start with John.
    $\displaystyle m+j+a=3j+j+j-2=43$
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    Re: Word Problem: Computing age comparing quantities

    Thanks Plato. Yes that is how I solved the problem after getting stuck on my initial approach. But I would still like to know why I am not having success comparing quantities using Monica.
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    Re: Word Problem: Computing age comparing quantities

    Quote Originally Posted by allyourbass2212 View Post
    . But I would still like to know why I am not having success comparing quantities using Monica.
    This $\displaystyle x + (\frac{1}{3x})+ (\frac{1}{3x} -2) = 43$ wrong.

    It should be $\displaystyle x + (\frac{x}{3})+ (\frac{x}{3} -2) = 43$
    Thanks from allyourbass2212
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    Re: Word Problem: Computing age comparing quantities

    Quote Originally Posted by allyourbass2212 View Post
    Thanks Plato. Yes that is how I solved the problem after getting stuck on my initial approach. But I would still like to know why I am not having success comparing quantities using Monica.
    monica, $\displaystyle x$

    john, $\displaystyle \frac{x}{3}$

    alicia, $\displaystyle \frac{x}{3}-2$

    $\displaystyle x + \frac{x}{3} + \frac{x}{3} - 2 = 43$

    $\displaystyle x + \frac{x}{3} + \frac{x}{3} = 45$

    $\displaystyle 3x + x + x = 135$

    $\displaystyle 5x = 135$

    $\displaystyle x = 27$, monica

    $\displaystyle \frac{x}{3} = 9$, john

    $\displaystyle \frac{x}{3} - 2 = 7$, alicia
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