The problem asks us to solve the rational equation $\frac{8}{z^2-9} = \frac{4}{z^2-3z}$ for $z$. We need to select the correct choice, either $z=$ some value, or "There is no solution".

1. Problem Description

The problem asks us to solve the rational equation $\frac{8}{z^2-9} = \frac{4}{z^2-3z}$ for $z$. We need to select the correct choice, either $z=$ some value, or "There is no solution".

2. Solution Steps

First, we factor the denominators:

$z^2 - 9 = (z-3)(z+3)$

$z^2 - 3z = z(z-3)$

Then, the given equation is

$\frac{8}{(z-3)(z+3)} = \frac{4}{z(z-3)}$

We multiply both sides by the least common denominator, $z(z-3)(z+3)$, to eliminate the fractions. Note that $z \ne 0, z \ne 3, z \ne -3$ since these values would make the denominators zero.

$8z = 4(z+3)$

$8z = 4z + 12$

$4z = 12$

$z = 3$

However, we noted earlier that $z \ne 3$, since $z=3$ would make some of the denominators zero. Therefore, $z=3$ is an extraneous solution and is not a valid solution.

Since we have no valid solutions, the correct answer is "There is no solution".

3. Final Answer

B. There is no solution