Antwoord:
Oplossing: #x <-7/2 of x> 1/2 #. In intervalnotatie: # (-oo, -7/2) uu (1/2, oo) #
Uitleg:
1) # 1/2 | 2x + 3 | -1> 1 of 1/2 | 2x + 3 | > 2 of | 2x + 3 | > 4 of 2x + 3> 4 of 2x> 1 of x> 1/2 # OF
2) # 1/2 | 2x + 3 | -1> 1 of 1/2 | 2x + 3 | > 2 of | 2x + 3 | > 4 of 2x + 3 <-4 of 2x <-7 of x <-7/2 #
Oplossing: #x <-7/2 of x> 1/2 #. In intervalnotatie:# (-oo, -7/2) uu (1/2, oo) #Ans
