(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

  5.   y=6; de=7; fg=8

  6.   x=4.2; jl=14.2; mn=13

step-by-step explanation:

when chords cross inside a circle, the product of the segment lengths is a constant.

5. dh×eh = fh×gh

  y = gh = dh×eh/fh = 3×4/2 . . . divide by the coefficient of gh

  y = 6

of course, the total chord length is the sum of the lengths of its segments.

  de = dh +eh = 3+4 = 7

  fg = fh +gh = 2+6 = 8

6. this problem works the same way as the previous one.

  x = lk = mk×nk/jk = 6×7/10 = 4.2

  jl = jk +lk = 10 +4.2 = 14.2

  mn = mk +nk = 6 +7 = 13

comment on chords and secants

you can think of the points h (problem 5) and k (problem 6) as points where the chords meet. a chord is part of a secant line, a line that intersects the circle in two points. one can also consider these points (h or k) to be the points where the respective secant lines meet.

then, the product that is a constant is the product of the distance from the meeting point (h or k) to one circle intersection with the distance from that meeting point to the other circle intersection. for problem 5, the constant is the product hd×he = hf×hg.

it turns out that this rule regarding the products of lengths to points of intersection is also true when the secants intersect outside the circle.

that is, you only need to remember one rule to work both kinds of problems.

could you reply to this with the question so i can ?