We are given two inequalities:

1. $-33 > -4x - 5$
2. $-4x - 5 \geq -49$

We need to solve both inequalities individually.

First inequality: $-33 > -4x - 5$

1. Add 5 to both sides: $-33 + 5 > -4x$ $-28 > -4x$

2. Divide both sides by $-4$. Remember, when dividing by a negative number, the inequality sign reverses: $\frac{-28}{-4} < x$ $7 < x$ or $x > 7$

Second inequality: $-4x - 5 \geq -49$

1. Add 5 to both sides: $-4x \geq -49 + 5$ $-4x \geq -44$

2. Divide both sides by $-4$, reversing the inequality: $\frac{-4x}{-4} \leq \frac{-44}{-4}$ $x \leq 11$

Combined solution:

The solution to the compound inequality is: $7 < x \leq 11$ or $x \in (7, 11]$

Would you like more details or have any questions?

Here are 5 related questions for practice:

1. Solve: $2x - 5 > 3$ and $3x + 2 \leq 17$.
2. Solve the inequality: $-5x + 7 \geq -3x - 5$.
3. What is the solution set for $-2x + 4 < 6$ or $3x - 5 \geq 7$?
4. Solve and graph: $5x - 10 < 0$ and $2x + 8 \geq 12$.
5. Solve for $x$: $4x - 3 > 2x + 7$.

Tip: Always reverse the inequality sign when multiplying or dividing both sides by a negative number.