The inequality given is:

$\frac{1}{x} < \frac{1}{x - 3}$

Let's solve this step by step:

Step 1: Analyze the domain

• For $\frac{1}{x}$, $x \neq 0$, because division by zero is undefined.
• For $\frac{1}{x - 3}$, $x \neq 3$, as this would also result in division by zero.

Thus, the domain of the inequality is $x \neq 0$ and $x \neq 3$.

Step 2: Cross-multiply

Since both sides are fractions, we can cross-multiply. However, we must be cautious about the signs of $x$ and $x-3$, because multiplying by a negative number flips the inequality. Let's break this into cases:

Case 1: $x > 3$

When $x > 3$, both $x$ and $x - 3$ are positive, so the inequality remains the same: $1 \cdot (x - 3) < 1 \cdot x$ Simplifying: $x - 3 < x$ This simplifies to: $-3 < 0$ This is always true, so for $x > 3$, the inequality holds.

Case 2: $0 < x < 3$

In this case, $x$ is positive, but $x - 3$ is negative. This means we need to flip the inequality when multiplying: $1 \cdot (x - 3) > 1 \cdot x$ Simplifying: $x - 3 > x$ This simplifies to: $-3 > 0$ This is never true, so there are no solutions in the interval $0 < x < 3$.

Case 3: $x < 0$

For $x < 0$, both $x$ and $x - 3$ are negative, so the inequality remains the same: $1 \cdot (x - 3) < 1 \cdot x$ Simplifying: $x - 3 < x$ This simplifies to: $-3 < 0$ This is always true, so for $x < 0$, the inequality holds.

Step 3: Conclusion

The inequality holds when $x < 0$ or $x > 3$. So the solution is:

$x \in (-\infty, 0) \cup (3, \infty)$

Would you like more details or have any questions?

Here are 5 related questions:

1. How do you handle inequalities when cross-multiplying?
2. What happens when multiplying or dividing by a negative number in inequalities?
3. How do you determine the domain of a rational function?
4. Can you graph the solution of a rational inequality?
5. How can we solve inequalities involving quadratic expressions?

Tip: Always be cautious when multiplying or dividing inequalities by variables, as their signs affect the direction of the inequality.