To solve the compound inequality $2x - 6 \leq -10 \, \text{or} \, 3x + 1 > 16$, we will solve each inequality separately and then combine the results.

1. Solving $2x - 6 \leq -10$:

$2x - 6 \leq -10$ Add 6 to both sides: $2x \leq -4$ Divide by 2: $x \leq -2$

2. Solving $3x + 1 > 16$:

$3x + 1 > 16$ Subtract 1 from both sides: $3x > 15$ Divide by 3: $x > 5$

Final solution:

The compound inequality uses the word "or," which means we combine both solution sets:

• From $2x - 6 \leq -10$, we get $x \leq -2$.
• From $3x + 1 > 16$, we get $x > 5$.

Thus, the solution is $x \leq -2$ or $x > 5$.

Would you like further clarification or details on any part of this process?

Here are some related questions for practice:

1. How would the solution change if the inequality was "and" instead of "or"?
2. Solve $4x - 7 \leq -3 \, \text{or} \, 2x + 5 > 11$.
3. What happens if both inequalities have the same solution range?
4. How do you graph $x \leq -2$ or $x > 5$ on a number line?
5. Solve the compound inequality $5x - 4 \geq 6 \, \text{and} \, x + 2 < 10$.

Tip: Always pay close attention to the conjunction "and" or "or" in compound inequalities, as they dictate how solution sets are combined.