The problem is to solve the equation $1 - \frac{9}{z-4} = \frac{-5}{2z-8}$ for $z$. We need to find the value of $z$ that satisfies the equation or determine if there is no solution.

1. Problem Description

The problem is to solve the equation $1 - \frac{9}{z-4} = \frac{-5}{2z-8}$ for $z$. We need to find the value of $z$ that satisfies the equation or determine if there is no solution.

2. Solution Steps

First, we can simplify the equation by noticing that $2z-8 = 2(z-4)$. The equation can then be rewritten as:

$1 - \frac{9}{z-4} = \frac{-5}{2(z-4)}$

Multiply both sides of the equation by $2(z-4)$ to eliminate the denominators. Note that this assumes $z \neq 4$.

$2(z-4) \left(1 - \frac{9}{z-4}\right) = 2(z-4) \left(\frac{-5}{2(z-4)}\right)$

$2(z-4) - 2(z-4)\frac{9}{z-4} = -5$

$2(z-4) - 18 = -5$

$2z - 8 - 18 = -5$

$2z - 26 = -5$

$2z = 21$

$z = \frac{21}{2}$

Now we check if $z = \frac{21}{2}$ is a valid solution. Since $z \neq 4$, we can substitute $z = \frac{21}{2}$ into the original equation:

$1 - \frac{9}{\frac{21}{2} - 4} = \frac{-5}{2(\frac{21}{2}) - 8}$

$1 - \frac{9}{\frac{21}{2} - \frac{8}{2}} = \frac{-5}{21 - 8}$

$1 - \frac{9}{\frac{13}{2}} = \frac{-5}{13}$

$1 - \frac{18}{13} = \frac{-5}{13}$

$\frac{13}{13} - \frac{18}{13} = \frac{-5}{13}$

$\frac{-5}{13} = \frac{-5}{13}$

The solution $z = \frac{21}{2}$ is valid.

3. Final Answer

$z = \frac{21}{2}$