\( k-3<-7 \) or \( k+5 \geq 8 \)

Let's solve the compound inequality: \[ k - 3 < -7 \quad \text{or} \quad k + 5 \geq 8 \] **1. Solve \( k - 3 < -7 \):** \[ k - 3 < -7 \\ \text{Add } 3 \text{ to both sides:} \\ k < -7 + 3 \\ k < -4 \] **2. Solve \( k + 5 \geq 8 \):** \[ k + 5 \geq 8 \\ \text{Subtract } 5 \text{ from both sides:} \\ k \geq 8 - 5 \\ k \geq 3 \] **Combined Solution:** The solution to the compound inequality is all real numbers \( k \) that satisfy either \( k < -4 \) **or** \( k \geq 3 \). **In Interval Notation:** \[ (-\infty, -4) \cup [3, \infty) \] **Graphical Representation:** ``` <---|---------|=========> -4 3 ``` - The open circle at -4 indicates that \( k \) can be any number less than -4. - The closed circle at 3 indicates that \( k \) can be 3 or any number greater than 3. **Final Answer:** All real numbers k with k < –4 or k ≥ 3. In interval notation:  (–∞, –4) ∪ [3, ∞)