We are given two inequalities:

$\frac{3x + 10}{2} > 2 \quad \text{and} \quad 3x - 6 < x - 2$

Let's solve them step by step.

First inequality:

$\frac{3x + 10}{2} > 2$ Multiply both sides by 2 to eliminate the fraction: $3x + 10 > 4$ Now subtract 10 from both sides: $3x > -6$ Divide by 3: $x > -2$

Second inequality:

$3x - 6 < x - 2$ First, subtract $x$ from both sides: $2x - 6 < -2$ Now add 6 to both sides: $2x < 4$ Finally, divide by 2: $x < 2$

Combine the two results:

From the first inequality, we have $x > -2$, and from the second inequality, we have $x < 2$.

Thus, the solution set is: $-2 < x < 2$

This corresponds to the interval $(-2, 2)$. Therefore, $A = -2$ and $B = 2$.

To find $A + B$:

$A + B = -2 + 2 = 0$

The value of $A + B$ is $\boxed{0}$.

Would you like further details, or do you have any questions?

Here are 5 related questions to explore:

1. What if one inequality was non-strict, like $\frac{3x + 10}{2} \geq 2$?
2. How does changing a coefficient in an inequality impact the solution range?
3. What happens if we add an additional condition, like $x \geq 0$?
4. How can interval notation help in representing solutions graphically?
5. How do absolute value inequalities differ from linear inequalities?

Tip: Always check if your inequality flips when dividing by negative numbers!