Question

which expression is equivalent to the following complex fraction?\\(\frac{\frac{-2}{x} + \frac{5}{y}}{\frac{3}{y} - \frac{2}{x}}\\)\\(\frac{3x - 2y}{-2y + 5x}\\)\\(\frac{-2y + 5x}{3x - 2y}\\)\\(\frac{(-2y + 5x)(3x - 2y)}{x^2 y^2}\\)\\(\frac{x^2 y^2}{(-2y + 5x)(3x - 2y)}\\)

Explanation:

Step1: Simplify the numerator

The numerator of the complex fraction is $\frac{-2}{x} + \frac{5}{y}$. To add these two fractions, we find a common denominator, which is $xy$. So we rewrite each fraction with the common denominator:
$\frac{-2}{x} = \frac{-2y}{xy}$ and $\frac{5}{y} = \frac{5x}{xy}$.
Adding them together: $\frac{-2y + 5x}{xy}$.

Step2: Simplify the denominator

The denominator of the complex fraction is $\frac{3}{y} - \frac{2}{x}$. Again, we find a common denominator of $xy$. Rewriting each fraction:
$\frac{3}{y} = \frac{3x}{xy}$ and $\frac{2}{x} = \frac{2y}{xy}$.
Subtracting them: $\frac{3x - 2y}{xy}$.

Step3: Divide the numerator by the denominator

Now we have the complex fraction as $\frac{\frac{-2y + 5x}{xy}}{\frac{3x - 2y}{xy}}$. Dividing by a fraction is the same as multiplying by its reciprocal, so this becomes $\frac{-2y + 5x}{xy} \times \frac{xy}{3x - 2y}$.
The $xy$ terms cancel out, leaving us with $\frac{-2y + 5x}{3x - 2y}$, which can also be written as $\frac{-2y + 5x}{3x - 2y}=\frac{5x - 2y}{3x - 2y}$ (rearranging the numerator terms) or $\frac{-2y + 5x}{3x - 2y}$ which matches the second option (when we write the numerator as $-2y + 5x$ and the denominator as $3x - 2y$).

Answer:

B. $\frac{-2y + 5x}{3x - 2y}$ (assuming the second option is labeled B, if the options were labeled A, B, C, D with the second option being B as per the order: first option A, second B, third C, fourth D)