How do you solve #5abs(2r + 3) -5 = 0#?

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Answer 1 · 2015-07-15T09:13:01

#r=-2,-1#

#5abs(2r+3)-5=0#

Add #5# to both sides of the equation.

#5abs(2r+3)=5#

Divide both sides by #5#.

#abs(2r+3)=1#

Separate the equation into one positive equation and one negative equation.

#2r+3=1# and #-(2r+3)=1#

Positive equation

#2r+3=1# =

#2r=1-3# =

#2r=-2#

Divide both sides by #2#.

#r=-1#

Negative equation

#-(2r+3)=1#

#-2r-3=1#

#-2r=1+3# =

#-2r=4#

Divide both sides by #-2#.

#r=4/(-2)# =

#r=-2#