+1859 need help with fractions plz
Given positive integers $x$ and $y$ such that $x\neq y$ and $\frac{1}{x} + \frac{1}{y} = \frac{1}{6}$, what is the smallest possible value for $x + y$?
+121 Let's work with the equation \(\frac{1}{x} + \frac{1}{y} = \frac{1}{6}\).
First, we can clear the fractions by multiplying both sides of the equation by \(6xy\):
\[6y + 6x = xy.\]
Rearrange the terms:
\[xy - 6x - 6y = 0.\]
Now, we can use Simon's Favorite Factoring Trick to factor this equation:
\[xy - 6x - 6y + 36 = 36.\]
\[(x - 6)(y - 6) = 36.\]
We are looking for two distinct positive integer solutions \(x\) and \(y\) such that \((x - 6)(y - 6) = 36\).
The pairs of factors of 36 are: \((1, 36), (2, 18), (3, 12), (4, 9), (6, 6)\).
If we let \(x - 6 = 1\) and \(y - 6 = 36\), we get \(x = 7\) and \(y = 42\), which doesn't satisfy \(x \neq y\).
If we let \(x - 6 = 2\) and \(y - 6 = 18\), we get \(x = 8\) and \(y = 24\).
So, the solution with the smallest possible \(x + y\) is when \(x = 8\) and \(y = 24\), which gives \(x + y = 8 + 24 = 32\).
Therefore, the smallest possible value for \(x + y\) is \(32\).
+121 Let's work with the equation \(\frac{1}{x} + \frac{1}{y} = \frac{1}{6}\).
First, we can clear the fractions by multiplying both sides of the equation by \(6xy\):
\[6y + 6x = xy.\]
Rearrange the terms:
\[xy - 6x - 6y = 0.\]
Now, we can use Simon's Favorite Factoring Trick to factor this equation:
\[xy - 6x - 6y + 36 = 36.\]
\[(x - 6)(y - 6) = 36.\]
We are looking for two distinct positive integer solutions \(x\) and \(y\) such that \((x - 6)(y - 6) = 36\).
The pairs of factors of 36 are: \((1, 36), (2, 18), (3, 12), (4, 9), (6, 6)\).
If we let \(x - 6 = 1\) and \(y - 6 = 36\), we get \(x = 7\) and \(y = 42\), which doesn't satisfy \(x \neq y\).
If we let \(x - 6 = 2\) and \(y - 6 = 18\), we get \(x = 8\) and \(y = 24\).
So, the solution with the smallest possible \(x + y\) is when \(x = 8\) and \(y = 24\), which gives \(x + y = 8 + 24 = 32\).
Therefore, the smallest possible value for \(x + y\) is \(32\).
