Let's solve each inequality separately.

1. Solving the inequality:

$-2x - 10 \geq -22$

Step 1: Add 10 to both sides. $-2x \geq -22 + 10$ $-2x \geq -12$

Step 2: Divide both sides by -2. Note: When dividing by a negative number, reverse the inequality sign. $x \leq \frac{-12}{-2}$ $x \leq 6$

2. Solving the inequality:

$-2x - 10 > -34$

Step 1: Add 10 to both sides. $-2x > -34 + 10$ $-2x > -24$

Step 2: Divide both sides by -2. Remember to reverse the inequality sign again. $x < \frac{-24}{-2}$ $x < 12$

Conclusion:

The solution to the system of inequalities is: $x \leq 6 \quad \text{and} \quad x < 12$ Since $x \leq 6$ is more restrictive, the final solution is: $x \leq 6$

Would you like further details or have any questions?

Here are some related questions to explore:

1. How would the solution change if the first inequality were strict ($>$)?
2. What happens when dividing inequalities by positive numbers?
3. How do you solve compound inequalities with “or” conditions instead of “and”?
4. How do inequality solutions change when squared terms are involved?
5. Can you graph the solution to this inequality on a number line?

Tip: Always remember to reverse the inequality sign when multiplying or dividing by a negative number!