The given expression is:

\[
\frac{\frac{3}{5}x - \frac{x}{15}}{\frac{1}{x} - \frac{1}{3}}
\]

Let's simplify this step by step.

### Step 1: Simplify the numerator
\[
\frac{3}{5}x - \frac{x}{15}
\]

To subtract these terms, we need a common denominator. The common denominator between 5 and 15 is 15. So, rewrite \(\frac{3}{5}x\) as:

\[
\frac{3}{5}x = \frac{9}{15}x
\]

Now subtract:
\[
\frac{9}{15}x - \frac{1}{15}x = \frac{8}{15}x
\]

So, the numerator simplifies to:
\[
\frac{8}{15}x
\]

### Step 2: Simplify the denominator
\[
\frac{1}{x} - \frac{1}{3}
\]

Again, we need a common denominator. The common denominator between \(x\) and 3 is \(3x\), so rewrite each fraction:

\[
\frac{1}{x} = \frac{3}{3x}, \quad \frac{1}{3} = \frac{x}{3x}
\]

Now subtract:
\[
\frac{3}{3x} - \frac{1}{3} = \frac{3 - x}{3x}
\]

So, the denominator simplifies to:
\[
\frac{3 - x}{3x}
\]

### Step 3: Simplify the entire expression
Now the expression becomes:
\[
\frac{\frac{8}{15}x}{\frac{3 - x}{3x}}
\]

This is a complex fraction, which can be simplified by multiplying by the reciprocal of the denominator:

\[
\frac{8}{15}x \times \frac{3x}{3 - x} = \frac{8}{15} \times \frac{3x^2}{3 - x}
\]

The final simplified expression is:

\[
\frac{8x^2}{5(3 - x)}
\]

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Let me know if you need more details or have any questions!

Here are 5 related questions:
1. How do you simplify more complex algebraic fractions?
2. What are common mistakes to avoid when simplifying fractions?
3. How do you find a common denominator between algebraic terms?
4. How can you factor algebraic expressions involving fractions?
5. How do you deal with complex fractions in more advanced algebra problems?

**Tip:** Always check for common factors before performing operations in complex fractions to simplify your work.

This detailed guide walks through the step-by-step simplification of the complex fraction 3/5x - x/15 / 1/x - 1/3. Key algebraic concepts such as finding a common denominator, simplifying the numerator and denominator, and working with complex fractions are covered. Suitable for students in Grades 9-12.

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