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

5 / (x-5) = x / (x-5) - 5/4.

Multiply through by 4 (x-5):

20=4x-5 (x-5).

20=4x-5x+25=25-x, x=5.

This is an extraneous solution because when x=5, the denominators become zero so division by zero occurs, and the result is meaningless.