Given positive integers x and y such that \(x \not = y\) and \(\frac {1}{x} + \frac {1}{y} = \frac {1}{12}\), what is the smallest possible positive value for \(x+y\)?
X=6
Y=6
That is not correct Nickolas.
\(\frac {1}{x} + \frac {1}{y} = \frac {1}{12}\\~\\ \frac{1}{x}<\frac{1}{12}\qquad \qquad \frac{1}{y}<\frac{1}{12} \\ x>12\qquad \qquad y>12 \\ so\\ x+y>24 \)
Correct Melody! In fact we have:
Thanks Alan,
I did think there might be a higher answer than mine.
I guess I should have said so :)
I also should have thought to use calculus. That never even crossed my mind
+239 
