Question 184875
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I don't blame you a bit for being stumped on this one.  I had to chew on it all afternoon.  But I did come up with two solutions.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{1}{x} + \frac{1}{y} = \frac{11}{60}]


Now if you add the two fractions containing the variables:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{x + y}{xy} = \frac{11}{60}]


From which you could derive two equations by equating the numerators and equating the denominators.  But there is a problem.  Setting the numerators equal to each other gives you:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x + y = 11]


And there are no two even numbers that will satisfy this equation.  This is where I was stumped for a while until it occured to me to make a little change to the equation, thus:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{1}{x} + \frac{1}{y} = \frac{22}{120}]


Now we can say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x + y = 22]


And


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ xy = 120]


Rearranging the first equation:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y = 22 - x]


And substituting in the second equation:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x(22 - x) = 120 \ \ \Rightarrow\ \ 22x - x^2 = 120 \ \ \Rightarrow\ \ x^2 - 22x + 120 = 0]


And this quadratic factors rather tidily:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x - 10)(x - 12) = 0]


Therefore:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x - 10 = 0 \ \ \Rightarrow\ \ x = 10]


or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x - 12 = 0 \ \ \Rightarrow\ \ x = 12]


And going back to:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y = 22 - x]


We can see that <i>y</i> is 12 when <i>x</i> is 10, or vice versa, so the two numbers are 10 and 12.


Check:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{1}{10} + \frac{1}{12} = \frac{6}{60} + \frac{5}{60} =\frac{11}{60}]


Done, right?  Not so fast, Sparky.  Like the guy on TV selling Ginzu knives says, "But wait...there's more!"


After obtaining that result, it occurred to me that if multiplying *[tex \Large \frac{11}{60}] by *[tex \Large \frac{2}{2}] resulted in a solution, perhaps there was some integer <i>k</i> > 1 such that multiplying *[tex \Large \frac{11}{60}] by *[tex \Large \frac{2k}{2k}] would result in a solution to this problem.


So I tried *[tex \Large \frac{44}{240}] but that yielded:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  x^2 - 44x + 240 = 0]


which has irrational roots, and is therefore not a solution (verification of this left as an exercise for the student)


Then I tried So I tried *[tex \Large \frac{66}{360}] yielding:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  x^2 - 66x + 360 = 0]


Which factors to:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x - 6)(x - 60) = 0]


Giving us 6 and 60 as the two numbers.


Check:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{1}{6} + \frac{1}{60} = \frac{10}{60} + \frac{1}{60} =\frac{11}{60}]


I actually tried about a hundred different values of <i>k</i> but only found the two I showed you that worked, so I'm pretty certain that these are the only two solutions to the problem as stated.


Thanks.  I had fun with this one.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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