jmc

Multiplying the first equation by $y$ and the second equation by $x,$ we get $$\begin{array}{rl}xy+1& =10y,\\ xy+1& =\frac{5}{12}x.\end{array}$$Then $10y=\frac{5}{12}x,$ so $y=\frac{1}{10}\cdot \frac{5}{12}x=\frac{1}{24}x.$ 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 $y=\frac{1}{6},\frac{1}{4},$ respectively, which, we can check, are valid solutions to the original system of equations.)