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 of $x$.
1. Given the equations:
   \[
   x + \frac{1}{y} = 10
   \]
   \[
   y + \frac{1}{x} = \frac{5}{12}
   \]

2. Multiply the two equations:
   \[
   \left( x + \frac{1}{y} \right) \left( y + \frac{1}{x} \right) = 10 \cdot \frac{5}{12}
   \]
   Expanding the left-hand side:
   \[
   xy + x \cdot \frac{1}{x} + \frac{1}{y} \cdot y + \frac{1}{y} \cdot \frac{1}{x} = \frac{50}{12}
   \]
   Simplifying:
   \[
   xy + 1 + 1 + \frac{1}{xy} = \frac{25}{6}
   \]
   \[
   xy + \frac{1}{xy} + 2 = \frac{25}{6}
   \]
   Subtract 2 from both sides:
   \[
   xy + \frac{1}{xy} = \frac{25}{6} - 2
   \]
   \[
   xy + \frac{1}{xy} = \frac{25}{6} - \frac{12}{6}
   \]
   \[
   xy + \frac{1}{xy} = \frac{13}{6}
   \]

3. Let \( z = xy \). Then:
   \[
   z + \frac{1}{z} = \frac{13}{6}
   \]
   Multiply both sides by \( z \):
   \[
   z^2 + 1 = \frac{13}{6}z
   \]
   Rearrange into a standard quadratic equation:
   \[
   6z^2 - 13z + 6 = 0
   \]

4. Solve the quadratic equation using the quadratic formula \( z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
   \[
   z = \frac{13 \pm \sqrt{169 - 144}}{12}
   \]
   \[
   z = \frac{13 \pm \sqrt{25}}{12}
   \]
   \[
   z = \frac{13 \pm 5}{12}
   \]
   \[
   z = \frac{18}{12} \quad \text{or} \quad z = \frac{8}{12}
   \]
   \[
   z = \frac{3}{2} \quad \text{or} \quad z = \frac{2}{3}
   \]

5. Consider \( z = \frac{3}{2} \):
   \[
   xy = \frac{3}{2}
   \]
   Substitute \( xy = \frac{3}{2} \) into the first equation:
   \[
   x + \frac{1}{y} = 10
   \]
   \[
   x + \frac{1}{\frac{3}{2x}} = 10
   \]
   \[
   x + \frac{2x}{3} = 10
   \]
   \[
   \frac{5x}{3} = 10
   \]
   \[
   5x = 30
   \]
   \[
   x = 6
   \]
   Then:
   \[
   y = \frac{3}{2x} = \frac{3}{2 \cdot 6} = \frac{1}{4}
   \]

6. Consider \( z = \frac{2}{3} \):
   \[
   xy = \frac{2}{3}
   \]
   Substitute \( xy = \frac{2}{3} \) into the first equation:
   \[
   x + \frac{1}{y} = 10
   \]
   \[
   x + \frac{1}{\frac{2}{3x}} = 10
   \]
   \[
   x + \frac{3x}{2} = 10
   \]
   \[
   \frac{5x}{2} = 10
   \]
   \[
   5x = 20
   \]
   \[
   x = 4
   \]
   Then:
   \[
   y = \frac{2}{3x} = \frac{2}{3 \cdot 4} = \frac{1}{6}
   \]

Therefore, the possible values of \( x \) are \( 4 \) and \( 6 \).

The final answer is \( \boxed{4, 6} \).