How do you solve #abs(x-1/3)-2=1?#?

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Answer 1 · 2018-04-22T08:14:36

#x=10/3# or #-8/3#

As #|x-1/3|-2=1#, we have

#|x-1/3|=1+2=3#

hence either #x-1/3=3# i.e. #x=3+1/3=10/3#

or #x-1/3=-3# i.e. #x=-3+1/3=-8/3#

Graphically this solution can be plotted as the intersection points of the functions on the LHS and the RHS as shown below :-

GeoGebra Classic app

Answer 2 · 2018-04-22T08:19:49

#x_1 = 10/3 #
#x_2 =-8/3#

Any algebraic problem with absolute values you always have to isolate the absolute value on one side of the equation.

In this case:

# |x-1/3| = 3#

remember that for absolute values you will have 2 answers because both positive and negative values have a positive absolute value:

#+-(x-1/3) = 3#

Solve for x:

#x_1 = 10/3 #
#x_2 =-8/3#

Graphically this solution can be plotted as the intersection points of the functions on the LHS and the RHS as shown below :-

GeoGebra Classic app