- Find a common denominator for the numerator: \(\frac{2}{x} - \frac{4}{y} = \frac{2y-4x}{xy}\).
- Find a common denominator for the denominator: \(\frac{-5}{y} + \frac{3}{x} = \frac{3y-5x}{xy}\).
- Divide the simplified numerator by the simplified denominator: \(\frac{2y-4x}{xy} \cdot \frac{xy}{3y-5x} = \frac{2y-4x}{3y-5x}\).
- Factor the numerator to reach the final form: \(\frac{2(y-2x)}{3y-5x}\).
- The correct option is \(\boxed{\frac{2(y-2x)}{3y-5x}}\).
Explanation
- Simplifying the Numerator
To simplify the given complex fraction \(\frac{\frac{2}{x}-\frac{4}{y}}{\frac{-5}{y}+\frac{3}{x}}\), we begin by addressing the numerator and the denominator separately. The numerator is \(\frac{2}{x} - \frac{4}{y}\). To combine these, we find a common denominator, which is \(xy\). Converting the terms gives \(\frac{2y}{xy} - \frac{4x}{xy} = \frac{2y-4x}{xy}\).
- Simplifying the Denominator
Next, we simplify the denominator \(\frac{-5}{y} + \frac{3}{x}\). Using the same common denominator \(xy\), we rewrite the terms: \(\frac{-5x}{xy} + \frac{3y}{xy} = \frac{3y-5x}{xy}\).
- Performing the Division
Now, we divide the simplified numerator by the simplified denominator: \[\frac{\frac{2y-4x}{xy}}{\frac{3y-5x}{xy}}\] Dividing by a fraction is equivalent to multiplying by its reciprocal: \[\frac{2y-4x}{xy} \times \frac{xy}{3y-5x}\] The common term \(xy\) in the numerator and denominator cancels out, leaving: \[\frac{2y-4x}{3y-5x}\]
- Factoring and Final Result
Finally, we factor the numerator \(2y-4x\) by taking out the common factor of \(2\), resulting in \(2(y-2x)\). Thus, the fully simplified expression is \(\frac{2(y-2x)}{3y-5x}\). Comparing this result to the given options, we confirm that it matches the second option.
Examples
Complex fractions appear frequently in physics and engineering when calculating total resistance in parallel circuits or when dealing with compound rates of change. For example, if you are calculating the equivalent resistance \(R_{eq}\) of two resistors in parallel, you might encounter expressions where you need to simplify the ratio of sums and differences of individual resistance values to understand how the components behave together. Mastering this algebraic simplification allows you to quickly reduce complex models to manageable forms for easier analysis.