We are given a system of inequalities:

1. \( 1 \geq -x - 1 \)
2. \( -x - 1 \geq -13 \)

### Solving the first inequality:

1. \( 1 \geq -x - 1 \)
   
   Add 1 to both sides to eliminate the constant term on the right:
   \[
   1 + 1 \geq -x - 1 + 1
   \]
   \[
   2 \geq -x
   \]

   Now, multiply both sides by -1 and reverse the inequality:
   \[
   -2 \leq x
   \]
   This simplifies to:
   \[
   x \geq -2
   \]

### Solving the second inequality:

2. \( -x - 1 \geq -13 \)

   Add 1 to both sides:
   \[
   -x - 1 + 1 \geq -13 + 1
   \]
   \[
   -x \geq -12
   \]

   Multiply both sides by -1 and reverse the inequality:
   \[
   x \leq 12
   \]

### Final Solution:

We now have two conditions:
\[
x \geq -2 \quad \text{and} \quad x \leq 12
\]

This means the solution to the system is:
\[
-2 \leq x \leq 12
\]

Thus, the solution set is \( x \in [-2, 12] \).

Would you like more details or have any questions?

Here are 5 related questions:

1. What happens if you reverse the inequalities' directions?
2. How can you graph the solution on a number line?
3. How would the solution change if the second inequality was strict, like \( -x - 1 > -13 \)?
4. Can you check the solution by plugging boundary values into the original inequalities?
5. How would the system change if a coefficient were added to the variable \( x \)?

**Tip:** When multiplying or dividing by a negative number, always remember to reverse the inequality sign!

Solve the system of inequalities: 1 ≥ -x - 1 and -x - 1 ≥ -13

Explore how to solve a system of inequalities 1 ≥ -x - 1 and -x - 1 ≥ -13 with a clear, step-by-step solution. This problem helps students understand key algebraic concepts like solving linear inequalities and handling negative coefficients. Ideal for Grades 7-9.

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