Let $x$ and $y$ be real numbers such that $\frac{\sin{x}}{\sin{y}} = 3$ and $\frac{\cos{x}}{\cos{y}} = \frac{1}{2}$. The value of $\frac{\sin{2x}}{\sin{2y}} + \frac{\cos{2x}}{\cos{2y}}$ can be expressed in the form $\frac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p + q$.
1. Given the equations:
   \[
   \frac{\sin{x}}{\sin{y}} = 3 \quad \text{and} \quad \frac{\cos{x}}{\cos{y}} = \frac{1}{2}
   \]
   We can express \(\sin{x}\) and \(\cos{x}\) in terms of \(\sin{y}\) and \(\cos{y}\):
   \[
   \sin{x} = 3 \sin{y} \quad \text{and} \quad \cos{x} = \frac{1}{2} \cos{y}
   \]

2. We need to find the value of \(\frac{\sin{2x}}{\sin{2y}} + \frac{\cos{2x}}{\cos{2y}}\). Using the double-angle identities:
   \[
   \sin{2x} = 2 \sin{x} \cos{x} \quad \text{and} \quad \sin{2y} = 2 \sin{y} \cos{y}
   \]
   \[
   \frac{\sin{2x}}{\sin{2y}} = \frac{2 \sin{x} \cos{x}}{2 \sin{y} \cos{y}} = \frac{\sin{x} \cos{x}}{\sin{y} \cos{y}}
   \]
   Substituting \(\sin{x} = 3 \sin{y}\) and \(\cos{x} = \frac{1}{2} \cos{y}\):
   \[
   \frac{\sin{2x}}{\sin{2y}} = \frac{3 \sin{y} \cdot \frac{1}{2} \cos{y}}{\sin{y} \cos{y}} = \frac{3}{2}
   \]

3. Next, we use the double-angle identity for cosine:
   \[
   \cos{2x} = 2 \cos^2{x} - 1 \quad \text{and} \quad \cos{2y} = 2 \cos^2{y} - 1
   \]
   We need to find \(\frac{\cos{2x}}{\cos{2y}}\):
   \[
   \frac{\cos{2x}}{\cos{2y}} = \frac{2 \cos^2{x} - 1}{2 \cos^2{y} - 1}
   \]
   Substituting \(\cos{x} = \frac{1}{2} \cos{y}\):
   \[
   \cos^2{x} = \left(\frac{1}{2} \cos{y}\right)^2 = \frac{1}{4} \cos^2{y}
   \]
   Therefore:
   \[
   \cos{2x} = 2 \left(\frac{1}{4} \cos^2{y}\right) - 1 = \frac{1}{2} \cos^2{y} - 1
   \]
   \[
   \cos{2y} = 2 \cos^2{y} - 1
   \]
   Thus:
   \[
   \frac{\cos{2x}}{\cos{2y}} = \frac{\frac{1}{2} \cos^2{y} - 1}{2 \cos^2{y} - 1}
   \]

4. To simplify \(\frac{\cos{2x}}{\cos{2y}}\), we need to find \(\cos^2{y}\). Using the Pythagorean identity:
   \[
   \sin^2{x} + \cos^2{x} = 1
   \]
   Substituting \(\sin{x} = 3 \sin{y}\) and \(\cos{x} = \frac{1}{2} \cos{y}\):
   \[
   (3 \sin{y})^2 + \left(\frac{1}{2} \cos{y}\right)^2 = 1
   \]
   \[
   9 \sin^2{y} + \frac{1}{4} \cos^2{y} = 1
   \]
   Using \(\sin^2{y} + \cos^2{y} = 1\):
   \[
   9 (1 - \cos^2{y}) + \frac{1}{4} \cos^2{y} = 1
   \]
   \[
   9 - 9 \cos^2{y} + \frac{1}{4} \cos^2{y} = 1
   \]
   \[
   9 - 1 = 9 \cos^2{y} - \frac{1}{4} \cos^2{y}
   \]
   \[
   8 = \frac{35}{4} \cos^2{y}
   \]
   \[
   \cos^2{y} = \frac{32}{35}
   \]

5. Substituting \(\cos^2{y} = \frac{32}{35}\) into the expression for \(\frac{\cos{2x}}{\cos{2y}}\):
   \[
   \cos{2x} = \frac{1}{2} \cdot \frac{32}{35} - 1 = \frac{16}{35} - 1 = \frac{16 - 35}{35} = \frac{-19}{35}
   \]
   \[
   \cos{2y} = 2 \cdot \frac{32}{35} - 1 = \frac{64}{35} - 1 = \frac{64 - 35}{35} = \frac{29}{35}
   \]
   \[
   \frac{\cos{2x}}{\cos{2y}} = \frac{\frac{-19}{35}}{\frac{29}{35}} = \frac{-19}{29}
   \]

6. Finally, we combine the results:
   \[
   \frac{\sin{2x}}{\sin{2y}} + \frac{\cos{2x}}{\cos{2y}} = \frac{3}{2} + \frac{-19}{29}
   \]
   Converting to a common denominator:
   \[
   \frac{3}{2} = \frac{3 \cdot 29}{2 \cdot 29} = \frac{87}{58}
   \]
   \[
   \frac{-19}{29} = \frac{-19 \cdot 2}{29 \cdot 2} = \frac{-38}{58}
   \]
   \[
   \frac{87}{58} + \frac{-38}{58} = \frac{87 - 38}{58} = \frac{49}{58}
   \]

The final answer is \(49 + 58 = \boxed{107}\)