Answer: B

Solving \(|x-5| = 2x\), we get two cases:

Case 1: \(x – 5 = 2x\)
\(x – 2x = 5\)
\(-x = 5\)
\(x = -5\)

Case 2: \(x – 5 = -2x\)
\(x + 2x = 5\)
\(3x = 5\)
\(x = \frac{5}{3}\)

Checking solutions in the original equation \(-2|x-5| = -4x\):

For \(x = -5\):
Left side: \(-2|-5 – 5| = -2| -10 | = -2 \times 10 = -20\)
Right side: \(-4 \times -5 = 20\)
\(-20 \neq 20\), so \(x = -5\) is not a solution.

For \(x = \frac{5}{3}\):
Left side: \(-2|\frac{5}{3} – 5| = -2|\frac{5}{3} – \frac{15}{3}| = -2|\frac{-10}{3}| = -2 \times \frac{10}{3} = -\frac{20}{3}\)
Right side: \(-4 \times \frac{5}{3} = -\frac{20}{3}\)
\(-\frac{20}{3} = -\frac{20}{3}\), so \(x = \frac{5}{3}\) is a solution.

Thus, the only valid solution is \(x = \frac{5}{3}\).