Let's solve the two inequalities step by step.

Inequality 1: $2x + 3 > 5x - 12$

Step 1: Isolate the variable $x$.

• Subtract $2x$ from both sides: $3 > 3x - 12$

• Add 12 to both sides: $15 > 3x$

• Divide by 3: $5 > x \quad \text{or} \quad x < 5$

Solution for inequality 1: $x < 5$

Inequality 2: $-\frac{1}{2}x + 4 \leq 0$

Step 1: Isolate the variable $x$.

• Subtract 4 from both sides: $-\frac{1}{2}x \leq -4$

• Multiply both sides by $-2$ (and remember to flip the inequality sign when multiplying by a negative): $x \geq 8$

Solution for inequality 2: $x \geq 8$

Final solutions:

1. $x < 5$
2. $x \geq 8$

Would you like more details on any step or have any questions?

Here are five related questions to consider:

1. How do you solve compound inequalities involving "and" or "or"?
2. What changes when multiplying or dividing inequalities by a negative number?
3. How would the solutions change if the inequalities involved fractions with $x$?
4. How can you represent the solutions of these inequalities on a number line?
5. What is the process for solving absolute value inequalities?

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