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We are tasked with solving the inequality:

\[
-12 \leq -2x - 3 < 0
\]

This is a compound inequality, and we will solve it step by step.

### Step 1: Add 3 to all parts of the inequality
We want to isolate the term involving \(x\), so we'll first eliminate the constant term \(-3\) by adding 3 to all parts of the inequality.

\[
-12 + 3 \leq -2x - 3 + 3 < 0 + 3
\]

This simplifies to:

\[
-9 \leq -2x < 3
\]

### Step 2: Divide by \(-2\) and flip the inequality signs
Since we are dividing by a negative number, we need to reverse the direction of the inequality signs. Dividing all parts by \(-2\):

\[
\frac{-9}{-2} \geq \frac{-2x}{-2} > \frac{3}{-2}
\]

This simplifies to:

\[
\frac{9}{2} \geq x > \frac{-3}{2}
\]

### Final Answer:
The solution is:

\[
\frac{-3}{2} < x \leq \frac{9}{2}
\]

This is the range of values that satisfy the inequality.

Would you like any further details or explanations?

Here are some related questions you might find helpful:
1. How do we solve a compound inequality with more complex terms?
2. What happens when we divide or multiply an inequality by a negative number?
3. Can compound inequalities have solutions that are infinite?
4. How do we graph the solution of a compound inequality on a number line?
5. How would the inequality change if the signs were reversed?

**Tip:** When solving inequalities, always remember to flip the inequality sign when multiplying or dividing by a negative number.

Learn how to solve the compound inequality -12 ≤ -2x - 3 < 0 with a detailed step-by-step solution. This problem involves key algebraic concepts such as inequality manipulations and the sign reversal rule when dividing by negative numbers. Suitable for students in Grades 9-10.

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