- Find a common denominator for the fractions in the numerator: \(\frac{2}{x-3} - \frac{3}{x} = \frac{9-x}{x(x-3)}\).
- Divide the new numerator by the denominator by multiplying by the reciprocal: \(\frac{9-x}{x(x-3)} \cdot \frac{x-3}{3}\).
- Cancel the common factor \((x-3)\) to obtain the simplified expression.
- The final simplified form is \(\boxed{\frac{9-x}{3x}}\).
Explanation
- Simplifying the Numerator
To simplify the expression \(\frac{\frac{2}{x-3} - \frac{3}{x}}{\frac{3}{x-3}}\), we first focus on the numerator. The numerator is the difference of two fractions: \(\frac{2}{x-3}\) and \(\frac{3}{x}\). To combine these, we find a common denominator, which is \(x(x-3)\). We write:
\[\frac{2}{x-3} - \frac{3}{x} = \frac{2x}{x(x-3)} - \frac{3(x-3)}{x(x-3)}\]
Now, combine the numerators over the common denominator:
\[\frac{2x - 3(x-3)}{x(x-3)} = \frac{2x - 3x + 9}{x(x-3)} = \frac{9-x}{x(x-3)}\]
Thus, the expression becomes \(\frac{\frac{9-x}{x(x-3)}}{\frac{3}{x-3}}\).
- Simplifying the Complex Fraction
Next, we simplify the complex fraction by multiplying the numerator by the reciprocal of the denominator. The denominator is \(\frac{3}{x-3}\), so its reciprocal is \(\frac{x-3}{3}\).
\[\frac{9-x}{x(x-3)} \cdot \frac{x-3}{3}\]
We observe that the factor \((x-3)\) appears in both the numerator and the denominator, so we can cancel it out (since \(x
eq 3\)):
\[\frac{9-x}{x \cdot 3} = \frac{9-x}{3x}\]
The expression is now in its simplest form.
- Final Conclusion
The simplified form of the given complex fraction is \(\frac{9-x}{3x}\). Therefore, the final answer is \(\boxed{\frac{9-x}{3x}}\).
Examples
Simplifying complex algebraic expressions is a fundamental skill in algebra, useful for solving equations, analyzing functions, and understanding mathematical relationships. In real-world engineering or physics problems, formulas for rates, concentrations, or ratios often result in complex fractions. Simplifying them into a clean form like \(\frac{9-x}{3x}\) makes calculations more efficient, reduces the likelihood of errors, and provides a clearer insight into how the result behaves as the input variable \(x\) changes. For example, if you are modeling a chemical reaction rate relative to volume, simplification allows you to quickly assess trends and limits.