Solve (3/x-3) =(x/x-/2) for x and determine if the solution is extraneous or not

3/(x-3)=x/(x-3)-3/2
6/2(x-3)=2x/2(x-3)-3(x-3)/2(x-3)
6=2x-3(x-3)
6=2x-3x+9
x=3

x=3 makes the denominator x-3 a zero, so the solution is extraneous, 

Distance=slopeslope is -2/4which equals-1/2the distance between these two points is -1/2