### Instruction: Given that $x$ and $y$ are nonzero real numbers such that $x+\frac{1}{y}=10$ and $y+\frac{1}{x}=\frac{5}{12},$ find all possible values for $x.$

(Enter your answer as a comma-separated list.).### Response: I notice that the two equations have a similar form, so I wonder if I can use substitution or elimination to solve for $x$ and $y.$
If I multiply the first equation by $y,$ I get $xy+1=10y.$
Similarly, if I multiply the second equation by $x,$ I get $xy+1=\frac{5}{12}x.$
These equations are equal, so I can set them equal to each other and solve for $x:$

$10y=\frac{5}{12}x$

$120y=5x$

$24y=x$
Now I can plug this into the first equation and solve for $y:$

$24y+\frac{1}{y}=10$

$24y^2+1=10y$

$24y^2-10y+1=0$
This is a quadratic equation, so I can use the quadratic formula to find the roots:

$y=\frac{10\pm\sqrt{100-4(24)(1)}}{2(24)}$

$y=\frac{10\pm\sqrt{4}}{48}$

$y=\frac{5\pm1}{24}$

$y=\frac{1}{4}$ or $y=\frac{1}{6}$