Question 1187966
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Let *[tex \Large x] represent the number of hours it takes the experienced crew to do the entire job by themselves.  Then *[tex \Large 2x] must represent the number of hours the inexperienced crew takes to do the entire job by themselves.


Since the experienced crew can do the entire job in *[tex \Large x] hours, they can do *[tex \Large \frac{1}{x}] of the job in one hour.  Likewise, the inexperienced crew can do *[tex \Large \frac{1}{2x}] of the job in one hour.


Then, working together, they can do *[tex \Large \frac{1}{x}\ +\ \frac{1}{2x}] of the job in one hour.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{1}{x}\ +\ \frac{1}{2x}\ =\ \frac{2x\,+\,x}{2x^2}\ =\ \frac{3x}{2x^2}\ =\ \frac{3}{2x}]


Since the two crews working together can do *[tex \Large \frac{3}{2x}] of the job in one hour, they can do the whole job in *[tex \Large \frac{2x}{3}] hours.  But we know that they take 6 hours working together, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{2x}{3}\ =\ 6]


Solve for *[tex \Large x]

																
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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From <https://www.algebra.com/cgi-bin/upload-illustration.mpl> 
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