https://file.aiquickdraw.com/mathbot/file/17269239085508iw9plfx.jpg

The inequality provided from the image is:

\[
\frac{3x - 1}{x} > x + 3
\]

Let's solve this step by step.

### Step 1: Rewrite the inequality

The inequality can be written as:

\[
\frac{3x - 1}{x} > x + 3
\]

Multiply both sides of the inequality by \(x\), but we must consider two cases:

1. \(x > 0\)
2. \(x < 0\)

We'll treat these two cases separately to ensure the inequality remains correct when multiplying by \(x\).

---

#### Case 1: \(x > 0\)

For \(x > 0\), multiplying both sides of the inequality by \(x\) does not change the direction of the inequality:

\[
3x - 1 > x(x + 3)
\]

Simplify the right-hand side:

\[
3x - 1 > x^2 + 3x
\]

Rearrange all terms to one side:

\[
0 > x^2 + 3x - 3x + 1
\]

This simplifies to:

\[
0 > x^2 + 1
\]

This inequality has no real solutions, because \(x^2 + 1\) is always positive for all real \(x\).

Thus, there are no solutions for \(x > 0\).

---

#### Case 2: \(x < 0\)

For \(x < 0\), multiplying both sides by \(x\) will reverse the inequality sign:

\[
3x - 1 < x(x + 3)
\]

Simplify the right-hand side:

\[
3x - 1 < x^2 + 3x
\]

Rearrange all terms:

\[
0 < x^2 + 1
\]

This is always true because \(x^2 + 1\) is always positive.

So, for \(x < 0\), the inequality holds for all negative \(x\).

### Conclusion

The solution to the inequality is:

\[
x < 0
\]

Would you like further clarification on any steps or more detailed explanation?

Here are 5 relative questions you might consider:

1. How do you solve inequalities involving absolute values?
2. Can inequalities with quadratic expressions have multiple solution ranges?
3. What are the conditions for flipping inequality signs?
4. How do you interpret solution sets on a number line?
5. What role do critical points play in solving inequalities?

**Tip:** Always check for the special cases where multiplying by a variable might reverse the inequality sign!

Solve the inequality \(\frac{3x - 1}{x} > x + 3\)

Learn how to solve the inequality \(\frac{3x - 1}{x} > x + 3\) with step-by-step solutions. The problem involves algebraic simplification, inequality transformation, and solving quadratic expressions. Suitable for Grades 9-12.

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