How do you solve #x+3=-abs(3x-1)#?

1 Answer
Alan P.
Apr 7, 2015

Consider two possibilities which are significant for the absolute value term

Possibility 1: #(x<=1/3)#
In this case #(3x-1) <= 0#
and the equation can be re-written
#x+3 = - (-(3x-1))#
#x+3 = 3x-1#
#-2x = -4#
#x= 2#
Note that this is an extraneous solution since it does not exist within the range of values for Possibility 1: #(x<=1/3)#

**Possibility 2: #(x>1/3)#
In this case #(3x-1) > 0#
and the equation can be re-written
#x+3 = -(3x -1)#
#4x = -2#
#x = -1/2#
Note that once again we have an extraneous solution sin it does not exist within the range of values for Possibility 2: (#x>1/3#)

To see why this happens consider a re-arrangement of the original equation into the form:
#x+3 +abs(3x-1) = 0#
The graph of the left side of this equation is shown below. Notice that it does not intersect the X-axis (that is it is never equal to #0#).
graph{x+3+abs(3x-1) [-16.01, 16.02, -8.01, 8]}

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