Hello from MrBillDoesMath!
Answer:
x >  8
Note: the quotient never equal 1. See Discussion below
Discussion:
When is (x + 2)/ (x - 8 ) >1?
As x + 2 = (x-8) + 8 + 2 = (x -8 ) + 10        (A)
(x + 2)/ (x - 8 ) >= 1               =>  using (A) above
( (x-8) + 10  ) /(x-8) > = 1       =>  do the division!
(x-8)/(x-8) + 10/(x-8) > = 1     => as (x-8)/(x-8) = 1
1 + 10/(x-8) >=1                     => subtract 1 from both sides
10/(x-8) >= 0                        => as 10 >= 0
x-8 >= 0                               => add 8 to both sides
x >=8
** Note the value of x = 8 is spurious as (x-8) is in the denominator of our computations and division by zero is not allowed. 
Also note that (x+2)/(x-8) can never equal 1. If it could, then
(x+2)/(x-8) =1    =>     multiply both sides by x-8
x+2 = x - 8       =>      subtract x form both sides
2 =  = -8           =>     Not possible
Thank you,
MrB