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
