!z!=a→z=a or z=-a; z=4x+3, a=9+2x
!4x+3!=9+2x
1) 4x+3=9+2x
Solving for x:
4x+3-3-2x=9+2x-3-2x
2x=6
2x/2=6/2
x=3
Checking for extraneous solution:
!4x+3!=9+2x
x=3→!4(3)+3!=9+2(3)
!12+3!=9+6
!15!=15
15=15 Ok, then x=3 is not a extraneous solution
2) 4x+3=-(9+2x)
Solving for x:
4x+3=-9-2x
4x+3-3+2x=-9-2x-3+2x
6x=-12
6x/6=-12/6
x=-2
Checking for extraneous solution:
!4x+3!=9+2x
x=-2→!4(-2)+3!=9+2(-2)
!-8+3!=9-4
!-5!=5
5=5 Ok, then x=-2 is not a extraneous solution
Answer:
x = -2 or 3
