How do you solve #| 3( x + 2) - 4 | = 1#?

Question

↳Redirected from "How do you evaluate #sin^-1 (-sqrt3/2)#?"

1918 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-09-01T14:58:23

#x =-1/3 and -1#

Start by distributing within the absolute value.

#|3x + 6 - 4| = 1#

#|3x + 2| = 1#

Case 1: The absolute value is positive

#3x + 2 = 1#

#3x = -1#

#x = -1/3#

Case 2: The absolute value is negative

#-(3x + 2) = 1#

#-3x - 2 = 1#

#-3x = 3#

#x = -1#

Hopefully this helps!