'drhong' has given you the answers but let me describe it a little different
Quote (JoeyMorgan619 @ 27 Jan 2014 04:00)
Ok so I have just a couple questions that I couldn't figure out. The answers are there, I just can't figure out how to get them.
http://i42.tinypic.com/tan3wj.png
I am pretty lost on this one altogether.
it is clear that when x is greater or equal zero the result will be positive
for the case that x is negative let's first multiply out:
x(2x+9) => 0 ie 2x^2+9x =>0 now x^2 will be positive while 9x will be negative for negative x
which means we have to find those negative x's with 2x^2 => 9x ie abs(-x) => 9/2 which gives the second segment
http://i42.tinypic.com/2vw61di.png
I was able to get 10≤x, but I don't know where the x<3 is coming from?
here you have (2x+1) / (x-3) <= 3 multiply both sides with (x-3) and you get 2x+1 <= 3x-9 ie 10 <= x which is one part, the one you got
you can directly see that the function has a singularity at 3, even without graphing the function so how does it look around +3?
x=+4 gives 9 while x=+2 gives -5 so the function switches at the singularity from being negative on the left side to positive on the right side
which implies that the function value is less than 3 when x is less than three (at least immediately to the left of x=3)
that it covers the whole range to the left can easily be seen with the function going towards +2 for very large negative x
(for very large x, the "+1" and "-3" stop to contribute anything meaningful)
http://i44.tinypic.com/2yugcad.png
I got the x≥-3/2, but I keep getting x≤-7/2 instead of 9/2?
8-abc(2x-3) => 2 ie 6 => abs (2x-3)
for positive x we get can rewrite that as 6 => 2x-3 ie 9 => 2x which means the inequality does not work for x>9/2
the other side you worked successfully you claim, so i leave it at thathope those comment help