Solving Inequalities

find the solution set for :

|2x+9| <= |x-6|

sidenotes - <= means greater or equal to. and " | | " means absolute value.

Thanks

Something like a<=|b| is equivalent to the inequalities -b<=a<=b. So, we can treat |2x+9| as a:
-x+6 <= |2x+9| <= x-6
Then, we can repeat the process for the |2x+9| <= x-6 side (we only need to do one side because the process will give the same inequalities each time since two absolute value graphs of lines whose "V" intersection point is on the x axis can only intersect at most two times - the maximum and minimum values we get from one side are these two intersections).
|2x+9| <= x-6
treat x-6 as "a" in a<=|b| this time:
-2x-9 >= x-6 >= 2x+9
so, solving each side, -2x-9 >= x-6 and x-6 >= 2x+9, gives -15<=x<=-1

bravo, thanx

actually, I have some questions...

Something like a<=|b| is equivalent to the inequalities -b<=a<=b. So, we can treat |2x+9| as a:
-x+6 <= |2x+9| <= x-6

i don't get that really, could you clarify/prove?

Secondly,

(we only need to do one side because the process will give the same inequalities each time since two absolute value graphs of lines whose "V" intersection point is on the x axis can only intersect at most two times - the maximum and minimum values we get from one side are these two intersections).

I really don't understand that... a visual perhaps?

Thanks for helping

"i don't get that really, could you clarify/prove?" -> Actually...my explanation really doesn't make sense. I got mixed up with where I put the absolute value.

You can think of the absolute value as meaning the "magnitude" of a number, or how far away from it is to zero. So, if |a|<=b (not a<=|b|), then -b<=a<=b (also, if |a|>=b, then a>=b or a<=-b). My answer is still correct, but I can re-solve it using my new explanation as a guide if it's unclear.

"I really don't understand that... a visual perhaps?" -> - each two distinct functions only intersects the other twice. Between these intersections is where one is greater than or equal to another.

I am very disappointed that you don't know how to embed that image, bfr.

There, I edited it

Way to not make it with a transparent background.

si, si, gracais muchas