Question: Consider the figure, where $\overrightarrow{RZ}$ is the angle bisector of $\angle QRS$, and $m\angle SRZ = 75^\circ$. If $m\angle QRZ = (-8x + 43)^\circ$, what is the value of $x$?
1. **State the problem:** We are given that $\overrightarrow{RZ}$ bisects $\angle QRS$, and $m\angle SRZ = 75^\circ$. We need to find $x$ given that $m\angle QRZ = (-8x + 43)^\circ$.
2. **Recall the angle bisector property:** Since $\overrightarrow{RZ}$ bisects $\angle QRS$, it divides the angle into two equal parts:
$$m\angle QRZ = m\angle SRZ$$
3. **Set up the equation:** Using the given values,
$$-8x + 43 = 75$$
4. **Solve for $x$:**
$$-8x + 43 = 75$$
Subtract 43 from both sides:
$$-8x + \cancel{43} - \cancel{43} = 75 - 43$$
$$-8x = 32$$
Divide both sides by $-8$:
$$\frac{-8x}{\cancel{-8}} = \frac{32}{\cancel{-8}}$$
$$x = -4$$
5. **Answer:** The value of $x$ is $-4$.