Let's solve the compound inequality:

$-3 < 3x + 3 \leq 10$

Step 1: Solve the first part of the compound inequality:

$-3 < 3x + 3$ Subtract 3 from both sides: $-3 - 3 < 3x$ $-6 < 3x$ Now, divide by 3: $\frac{-6}{3} < x$ $-2 < x$

Step 2: Solve the second part of the compound inequality:

$3x + 3 \leq 10$ Subtract 3 from both sides: $3x \leq 7$ Now, divide by 3: $x \leq \frac{7}{3}$

Step 3: Combine the two inequalities:

$-2 < x \leq \frac{7}{3}$

Step 4: Write the solution in interval notation:

The solution is the interval $(-2, \frac{7}{3}]$.

Let me know if you need more details or have any questions!

Here are some related questions for practice:

1. How do you solve inequalities with multiple variables?
2. What are the rules for dividing inequalities by negative numbers?
3. Can you explain the concept of interval notation?
4. How does the solution change if the inequality symbol is reversed?
5. How can compound inequalities be graphed on a number line?

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