|x+3|-3greater than or equal to 1

|x+3|-3 > 1

Two case by replacing equal sign, a related equation.

Case one x+3 -3-1 = 0

X-1 =0

X = 1

Case two -(x+3)-3-1 =0

-x -3-4 =0

-x-7 =0

-x = 7

x = -7

using a number line analysis: Check each interval (- infinity, -7], [-7, 1] and [1, + infinity) to determine which interval is greater than zero.

Test values are -2, 0, and -8

1st interval (- infinity, -7], test value -8

|-8+3|-3-1 > 0

|-7|-4 > 0

7-4 > 0

3 > 0 True,

2nd interval [-7, 1], test value is -2

|-2+3|-3-1 > 0

1-4 > 0

-3 > 0 false, doesn’t met

2nd interval (1, + infinity), test value is 2

|2+3|-3-1 > 0

5-4 > 0

1> 0 True

Therefore, the solution are (- infinity, -7] union [1, + infinity)