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.