Question 1166045
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The older can do the job in *[tex \Large x] hours by himself, therefore he can do *[tex \Large \frac{1}{x}] of the job in 1 hour.  The younger brother takes two hours longer, so he can do the job in *[tex \Large x\ +\ 2] hours by himself, and therefore can do *[tex \Large \frac{1}{x\,+\,2}] of the job in one hour.  Working together, they can do the job in 6 hours, so they can do *[tex \Large \frac{1}{6}] of an hour.  Taking this altogether:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{1}{x}\ +\ \frac{1}{x\,+\,2}\ =\ \frac{1}{6}]


Solve for *[tex \Large x] and then calculate *[tex \Large x\ +\ 2].  


By the way, the answers are irrational, but you are only off by about 2 and a half minutes on the 6 hour total if you round *[tex \Large x] to the nearest whole number.


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