Question 529683
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Your assignment of variables was spot on, though you could just as easily used *[tex \Large x] for Bob and *[tex \Large x\ +\ 4] for Dave.  Either way it will all come out in the wash.  Let's use your assignments, but first take a look at a general discussion of "Working Together" problems.


If A can do a job in <i>x</i> time periods, then A can do *[tex \Large \frac{1}{x}] of the job in 1 time period.  Likewise, if B can do the same job in <i>y</i> time periods, then B can do *[tex \Large \frac{1}{y}] of the job in 1 time period.


So, working together, they can do


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \frac{1}{x}\ +\ \frac{1}{y}\ =\ \frac{x\ +\ y}{xy} ]


of the job in 1 time period.


Therefore, they can do the whole job in:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \frac{1}{\frac{x + y}{xy}}\ =\ \frac{xy}{x\ +\ y}]


time periods.


For this particular problem we need to work backwards through the discussion above.  We know that working together they get the job done in 16 time periods, so working together they must be able to accomplish *[tex \Large \frac{1}{16}] of the job in one time period, i.e. 1 hour.


Using your variable assignments, we can say that Dave can do *[tex \Large \frac{1}{x}] of the job in one time period, and Bob can do *[tex \Large \frac{1}{x\ -\ 4}] of the job in one time period, so working together they do:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{1}{x}\ +\ \frac{1}{x\ -\ 4}]


of the job in one time period.  But we have already established that they can do *[tex \Large \frac{1}{16}] of the job in one time period.  Hence:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{1}{x}\ +\ \frac{1}{x\ -\ 4}\ =\ \frac{1}{16}]


Just solve for *[tex \Large x], then calculate *[tex \Large x\ -\ 4]


Note that this equation does, indeed, reduce to a quadratic equation.  It will not factor so you will need to use the quadratic formula.  Also note that one of your roots will be less than 4, meaning that once you subtract 4 to get Bob's time, you get a negative number.  To avoid having to deal with the absurd notion that Bob can finish painting a house some number of hours before he starts, discard the smaller root as extraneous.
  


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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