# 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?
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 • Jul 9th 2012, 03:15 PM Plato Re: Word Problem: Computing age comparing quantities Quote: Originally Posted by allyourbass2212 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$• Jul 9th 2012, 03:23 PM allyourbass2212 Re: Word Problem: Computing age comparing quantities Thanks Plato. Yes that is how I solved the problem after getting stuck on my initial approach. But I would still like to know why I am not having success comparing quantities using Monica. • Jul 9th 2012, 03:29 PM Plato Re: Word Problem: Computing age comparing quantities Quote: Originally Posted by allyourbass2212 . But I would still like to know why I am not having success comparing quantities using Monica. This$\displaystyle x + (\frac{1}{3x})+ (\frac{1}{3x} -2) = 43$wrong. It should be$\displaystyle x + (\frac{x}{3})+ (\frac{x}{3} -2) = 43$• Jul 9th 2012, 03:32 PM skeeter Re: Word Problem: Computing age comparing quantities Quote: Originally Posted by allyourbass2212 Thanks Plato. Yes that is how I solved the problem after getting stuck on my initial approach. But I would still like to know why I am not having success comparing quantities using Monica. monica,$\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