Let's work through both of the equations step by step.

First Equation:

$3(u - 2) - 5u = -2(u + 3)$

1. Expand both sides of the equation: $3u - 6 - 5u = -2u - 6$

2. Combine like terms: $(3u - 5u) - 6 = -2u - 6$ This simplifies to: $-2u - 6 = -2u - 6$

3. Notice that both sides of the equation are exactly the same, which means all real numbers are solutions for $u$.

Answer for the first equation: All real numbers are solutions.

Second Equation:

$-2(w + 1) + 5w = 3(w + 1) - 2$

1. Expand both sides of the equation: $-2w - 2 + 5w = 3w + 3 - 2$

2. Combine like terms: $(-2w + 5w) - 2 = 3w + 1$ This simplifies to: $3w - 2 = 3w + 1$

3. Subtract $3w$ from both sides: $-2 = 1$

Since this is a contradiction, there is no solution for $w$.

Answer for the second equation: No solution.

Would you like more details or have any questions?

Here are five related questions to expand your understanding:

1. How do you identify when an equation has no solution?
2. What are the steps to solve equations involving distribution?
3. How can we tell when all real numbers are solutions in an equation?
4. What happens if we encounter terms like $w^2$ instead of linear terms?
5. Can you solve systems of linear equations with two variables?

Tip: Always combine like terms and simplify each side of an equation before attempting to solve it.