Question 1199370
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The jump from
|x+2| < 5
to
-5 < x + 2 < 5
is valid.


The same can be said when you went from
 -5 < x + 2 < 5
to
-7 < x < 3


Then, as the tutor @greenestamps has pointed out, you subtract 2 from all sides to get x-2 in the middle.
-7 < x < 3
-7-2 < x-2 < 3-2
-9 < x-2 < 1


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Or you can subtract 4 from all sides after the absolute value bars have been removed.


Why 4? 
Because we want to go from x+2 to x-2
The gap from +2 to -2 on the number line is 4 units.


I.e.
x+2 to (x+2)-4 to x-2


So,
-5 < x + 2 < 5
-5-4 < (x + 2)-4 < 5-4
-9 < x-2 < 1
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