OG 12 PS 130

On the number line you can see that -5 <= x <= 3

you can use back solving here, from choice E.

|X+1|<=4

x + 1 <= 4

x <= 3

-|X+1|<=4

-x - 1 <= 4

-x <= 5

x >= -5

so it fits perfectly.

-5 <= x <= 3

Also, you can eliminate A and B b/c they are not closed intervals.

where is the number line??

I got both B and E fit into this. But why can't we use anything other than closed intervals? Even B gives the same output right?

B is
X>= -5 to X <= 5 (incorrect since it includes region from 3 to 5 as well)

E is
X+1 <= 4 ; X+1 >= -4
i.e. X <=3; x>=-5
this satisfies the whole region

Hope its clear now.

reasoning way:
|x-a| is the distance from any point x to point a.
just look at the number line and see the options:

A. |x| <= 3 distance from any point to 0 is less than or equal to 3-wrong
B. |X| <= 5 distance from any point to 0 is less than or equal to 5-wrong
C. |X-2| <= 3 distance from any point to 2 is less than or equal to 3-wrong
D. |X-1|<= 4 distance from any point to 1 is less than or equal to 4-wrong
E. |X+1|<=4 distance from any point to -1 is less than or equal to 4-right