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Vlada [557]
3 years ago
9

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

Mathematics
1 answer:
Ann [662]3 years ago
3 0

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

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