Solve 5 over x minus 5 equals the quantity of x over x minus 5, minus five fourths for x and determine if the solution is extraneous or not. (1 point)

Assuming the equation is:
5/(x-5) = x/(x-5) - 5x/4
We first multiply by the LCD: 4(x-5)
20 = 4x - 5x(x-5)
20 = 4x - 5x^2 + 25x
5x^2 - 29x + 20 = 0
(5x - 4)(x - 5) = 0
x = 4/5, 5
Substituting x = 5 gives denominators of 0, which is extraneous.
Substituting x = 4/5 gives a valid equation, so this is the only correct solution.