Let's go through each inequality and find the solution set.

a. $3x - 1 > 4$

1. Add 1 to both sides: $3x > 5$

2. Divide by 3: $x > \frac{5}{3}$

Solution: $x > \frac{5}{3}$.

b. $\frac{3}{x + 2} \geq 1$

1. Multiply both sides by $x + 2$ (considering the sign change if $x + 2$ is negative):

   • If $x + 2 > 0$ (i.e., $x > -2$): $3 \geq x + 2 \quad \Rightarrow \quad 3 - 2 \geq x \quad \Rightarrow \quad 1 \geq x$ This gives $x \leq 1$ for $x > -2$.

   • If $x + 2 < 0$ (i.e., $x < -2$), multiplying by a negative changes the inequality direction: 3 \leq x + 2 \quad \Rightarrow \quad \text{No solution since 3 cannot be less than a negative number.}

Solution: $-2 < x \leq 1$.

c. $3 < 4 - 2x \leq 7$

This is a compound inequality that can be split into two parts.

1. Solve $3 < 4 - 2x$: $3 - 4 < -2x \quad \Rightarrow \quad -1 < -2x \quad \Rightarrow \quad \frac{1}{2} > x$ So, $x < \frac{1}{2}$.

2. Solve $4 - 2x \leq 7$: $-2x \leq 7 - 4 \quad \Rightarrow \quad -2x \leq 3 \quad \Rightarrow \quad x \geq -\frac{3}{2}$

Solution: $-\frac{3}{2} \leq x < \frac{1}{2}$.

d. $x^2 - 5x + 6 < 2$

1. Rearrange the inequality: $x^2 - 5x + 6 - 2 < 0 \quad \Rightarrow \quad x^2 - 5x + 4 < 0$

2. Factor the quadratic: $(x - 4)(x - 1) < 0$

3. Use the sign analysis: The solution lies between the roots, so $1 < x < 4$.

Solution: $1 < x < 4$.

e. $|x - 2| < |x + 5|$

For absolute values, we need to consider different cases:

1. Case 1: $x \geq 2$

   For $x \geq 2$, the absolute values simplify to: $x - 2 < x + 5$ This inequality holds for all $x \geq 2$.

2. Case 2: $-5 \leq x < 2$

   For $-5 \leq x < 2$, we solve: $2 - x < x + 5$ Rearranging gives: $7 < 2x \quad \Rightarrow \quad x > \frac{7}{2}$ But $x$ is less than 2, so this case gives no solution.

3. Case 3: $x < -5$

   For $x < -5$, the absolute values become: $2 - x < -x - 5$ Rearranging gives no solution.

Solution: $x > -\frac{7}{2}$.

Let me know if you'd like further details or explanations. Here are 5 related questions to deepen your understanding:

1. How do you handle inequalities involving absolute values in general?
2. What is the importance of considering different cases in absolute value problems?
3. How do you solve quadratic inequalities using factoring?
4. What are the key steps in solving compound inequalities?
5. How does sign change affect inequality solutions?

Tip: When solving inequalities, always consider the sign of the terms you are dividing or multiplying by, as this may affect the inequality direction!