Three real numbers $x$, $y$, and $z$ are such that $(x+4)/2=(y+9)/(z-3)=(x+5)/(z-5)$.  Determine the value of $x/y$.
Given the equations:
\[
\frac{x+4}{2} = \frac{y+9}{z-3} = \frac{x+5}{z-5}
\]
Let this common ratio be \( k \). Therefore, we can write:
\[
\frac{x+4}{2} = k \implies x + 4 = 2k \implies x = 2k - 4
\]
\[
\frac{y+9}{z-3} = k \implies y + 9 = k(z-3) \implies y = kz - 3k - 9
\]
\[
\frac{x+5}{z-5} = k \implies x + 5 = k(z-5) \implies x = kz - 5k - 5
\]

Now, we have two expressions for \( x \):
\[
x = 2k - 4
\]
\[
x = kz - 5k - 5
\]

Equating these two expressions for \( x \):
\[
2k - 4 = kz - 5k - 5
\]
Rearrange to solve for \( z \):
\[
2k - 4 + 5k + 5 = kz
\]
\[
7k + 1 = kz
\]
\[
z = \frac{7k + 1}{k}
\]

Next, substitute \( z \) back into the expression for \( y \):
\[
y = k \left( \frac{7k + 1}{k} \right) - 3k - 9
\]
\[
y = 7k + 1 - 3k - 9
\]
\[
y = 4k - 8
\]

Now, we have:
\[
x = 2k - 4
\]
\[
y = 4k - 8
\]

We need to find \( \frac{x}{y} \):
\[
\frac{x}{y} = \frac{2k - 4}{4k - 8}
\]
Factor out the common terms:
\[
\frac{x}{y} = \frac{2(k - 2)}{4(k - 2)} = \frac{2}{4} = \frac{1}{2}
\]

The final answer is \(\boxed{\frac{1}{2}}\).