Let's solve the inequality:

$\frac{1}{x} + 3 > 4$

Step 1: Subtract 3 from both sides:

$\frac{1}{x} > 4 - 3$

$\frac{1}{x} > 1$

Step 2: Invert the inequality by multiplying both sides by $x$. But we need to consider two cases:

• If $x > 0$, the inequality stays the same.
• If $x < 0$, the inequality flips.

Case 1: $x > 0$

If $x > 0$, we can multiply both sides by $x$ without flipping the inequality:

$1 > x$

So, $x < 1$ for $x > 0$.

Case 2: $x < 0$

If $x < 0$, multiplying both sides by $x$ flips the inequality:

$1 < x$

This is a contradiction, because we assumed $x < 0$, so there are no solutions in this case.

Step 3: Combine the results:

The only valid solution is for $0 < x < 1$, so the solution set is:

$S = (0, 1)$

This matches the solution you provided.

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Would you like further details or clarification on any steps? Here are some related questions to explore:

1. What happens if we multiply or divide inequalities by negative numbers?
2. How can we solve rational inequalities with variables in both the numerator and denominator?
3. How do we handle inequalities involving absolute values?
4. What are the general rules for solving inequalities?
5. How does the graph of $\frac{1}{x}$ help visualize the solution to this inequality?

Tip: When solving inequalities, always pay close attention to whether you are multiplying or dividing by negative numbers, as this can flip the inequality.