Answer
$w=-18$
Work Step by Step
Given \begin{equation}
\frac{4}{w^2+2 w-24}=\frac{-0.5 w}{w^2-13 w+36}.
\end{equation}
Find the zeros of the denominators by factoring.
\begin{equation}
\begin{aligned}
w^2-13 w+36&=0\\
w^2-9 w-4 w+36& =0\\
w(w-9)-4(w-9)&=0 \\
(w-9)(w-4)& =0\\
\implies w= 9 && w= 4
\end{aligned}
\end{equation} \begin{equation}
\begin{aligned}
w^2+2 w-24&=0\\
w^2+6 w-4 w-24& =0\\
w(w+6)-4(w+6)&=0 \\
(w+6)(w-4)& =0\\
\implies w= -6 && w= 4.
\end{aligned}
\end{equation} Rewrite the original equation and cancel out like terms from both sides.
\begin{equation}
\begin{aligned}
\frac{4}{w^2+2 w-24}&=\frac{-0.5 w}{w^2-13 w+36}\\
\frac{4}{(w+6)(w-4)}&=\frac{-0.5 w}{(w-9)(w-4)}\\
\frac{4}{w+6}&=\frac{-0.5 w}{w-9}.
\end{aligned}
\end{equation} Cross multiply and solve for $w$: \begin{equation}
\begin{aligned}
\frac{4}{w+6}&=\frac{-0.5 w}{w-9}\\
4(w-9)&=-(w+6) \cdot 0.5 w\\
4(w-9)+(w+6) \cdot 0.5 w&=0\\
4w-36+0.5w^2+3w&=0\\
0.5w^2+7w-36&=0\\
w^2+14w-72&= 0\\
w^2+18w-4w-72&= 0\\
w(w+18)-4(w+18)&=0\\
\implies w= -18 \quad w= 4.
\end{aligned}
\end{equation} Check \begin{equation}
\begin{aligned}
\frac{4}{(-18)^2+2 \cdot (-18)-24}& \stackrel{?}{=} \frac{-0.5 \cdot (-18)}{(-18)^2-13 \cdot (-18)+36}\\
\frac{1}{66} & =\frac{1}{66}\checkmark\\
\frac{4}{(4)^2+2 \cdot (4)-24}& \stackrel{?}{=} \frac{-0.5 \cdot (4)}{(4)^2-13 \cdot (4)+36}\\
\frac{1}{2} & =\frac{-2}{0}\quad\textbf{False}\\
\end{aligned}
\end{equation} Note that $4$ is a zero of the denominators and can not be excepted as an answer.
The solution is $w=-18$.
