We are given the inequality:

$3x + 4 > 1 \quad \text{or} \quad x - 5 \leq 3 - 3x$

We will solve each part separately.

1. Solve $3x + 4 > 1$:

Subtract 4 from both sides: $3x > 1 - 4$ $3x > -3$ Divide by 3: $x > -1$

Thus, the solution to this part is: $x > -1$

2. Solve $x - 5 \leq 3 - 3x$:

Add $3x$ to both sides: $x + 3x - 5 \leq 3$ $4x - 5 \leq 3$ Add 5 to both sides: $4x \leq 3 + 5$ $4x \leq 8$ Divide by 4: $x \leq 2$

Thus, the solution to this part is: $x \leq 2$

Combine the results:

We now have two conditions: $x > -1 \quad \text{or} \quad x \leq 2$ Since the union of these two conditions covers all real numbers (because $x > -1$ already includes all values greater than $-1$ and the other inequality handles values less than or equal to 2), the solution is:

$\boxed{(-1, \infty)}$

Conclusion:

The solution in interval notation is $(-1, \infty)$.

Do you want further clarification on any steps, or have questions?
Here are five related questions to expand your understanding:

1. How do you solve inequalities with "and" instead of "or"?
2. How do you graph solutions of inequalities on a number line?
3. Can you combine multiple inequalities into a compound inequality?
4. What is the difference between strict inequalities and non-strict inequalities?
5. How do you handle inequalities involving absolute values?

Tip: When solving compound inequalities, always pay attention to whether the conditions are connected by "and" or "or", as this changes the solution set.