\(\frac{5x-4y}{2x-y} \times \frac{y-2x}{8y-1x}\)

\(\frac{5x-4y}{8y-10x}\)

\(\frac{x-y}{2y-2x}\)

Which is basically \(\frac{1}{2}\)

Regards

When we check the answer choices (ALWAYS check the answer choices before deciding on an approach!), we see that none of the answer choices include the variables x and y.
This tells us that all of the variables must cancel out.

Given this, let's assign some values to x and y, and then evaluate the given expression to see what we get.

If \(x = 1\), and \(y = 0\), we get: \(\frac{\frac{5x-4y}{2x-y}}{\frac{3y}{y-2x} + 5}=\frac{\frac{5(1)-4(0)}{2(1)-(0)}}{\frac{3(0)}{(0)-2(1)} + 5}\)

\(=\frac{(\frac{5}{2})}{\frac{0}{-2} + 5}\)

\(=\frac{(\frac{5}{2})}{5}\)

\(=\frac{1}{2}\)

Answer: A

Cheers,
Brent

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