(x+2)/(x-8) greater than and equal to 1

Hello from MrBillDoesMath!

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

the cat is blue because the probability of extension towards the prefix of the diameter of the circle could only mathematically equal to the sum of the roots of the carrots from   your garden.

step-by-step explanation: