Too bad you haven't shared illustrations of the possible solutions.
Starting with 4|x+3|>8, div. both sides by 4: |x+3|>2
Case 1: x+3 is already positive: then x+3>2, or x > -1
Case 2: x+3 is negative: Then -(x+3)>2, or -x - 3 > 2, or -5>x or x<-5
Draw a number line representing x values. Place empty circles at x = -5 and x = -1. Draw a vector from the circle at x = -1, to the right of x = -1. Draw a vector from the circle at x = - 5, to the left of x = -5. Note that x values between x = -5 and x = -1 are not solutions.
