How do you solve #|3x + 6| >15#?

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Answer 1 · 2018-03-28T12:55:00

#(-oo,-7)uu(3,oo)#

The absolute value is defined as:

#|a| = acolor(white)(888888)# if and only if #color(white)(88888)a>=0#

#|a| = -acolor(white)(.888) # if and only if #color(white)(888888)a<0#

Because of this we need to solve both:

#3x+6>15# and #-(3x+6)>15#

I

#3x+6>15#

#3x>9#

#x>3#

II

#-(3x+6)>15#

#3x+6<-15#

#3x<-21#

#x<-7#

This gives us:

#-oo < x< -7#

and

#3 < x < oo#

We can express this as a union of intervals:

#(-oo,-7)uu(3,oo)#