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 $x.$

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Multiplying the first equation by $y$ and the second equation by $x,$ we get \[\begin{aligned} xy+1 &= 10y, \\ xy + 1 &= \tfrac{5}{12} x. \end{aligned}\]Then $10y = \tfrac{5}{12}x,$ so $y = \tfrac{1}{10} \cdot \tfrac{5}{12} x = \tfrac{1}{24}x.$ Substituting into the first equation, we get \[x + \frac{1}{\frac{1}{24}x} = 10,\]or $x + \frac{24}{x} = 10,$ which rearranges to the quadratic $x^2 - 10x + 24 = 0.$ This quadratic factors as $(x-4)(x-6) = 0,$ so the possible values for $x$ are $\boxed{4, 6}.$ (These give corresponding $y$-values $y = \tfrac16, \tfrac14,$ respectively, which, we can check, are valid solutions to the original system of equations.)