How do you solve #|2x - 1|= 3#?

Given: #|2x - 1|= 3#

Please observe that #f(x) = 2x-1#; substitute this into the piecewise definition and set it equal to the right side (3):

#|2x-1| = {(2x-1; 2x-1 >= 0),(-(2x-1); 2x-1 < 0) :} = 3#

Separate the separate the pieces into two equations:

#2x - 1= 3; 2x-1 >=0 # and #-(2x - 1)= 3; 2x-1 <0 #

The inequalities that describe the domain restrictions probably do not make much sense to you, in their current form, so let's take a moment to simplify them:

#2x - 1= 3; 2x >=1 # and #-(2x - 1)= 3; 2x <1 #

#2x - 1= 3; x >=1/2 # and #-(2x - 1)= 3; x <1/2#

It is important to do this step so that you can see if an answer falls outside to the domain. For example, if the answer to the equation on the left was found to be #x = 1/4#, then you would need to discard that answer.

Now, we shall solve the equations:

#2x - 1= 3; x >=1/2 # and #2x - 1= -3; x <1/2#

#2x = 4; x >=1/2 # and #2x = -2; x <1/2#

#x = 2; x >=1/2 # and #x = -1; x <1/2#

Please observe that both answers fall within their respective domain restriction, therefore, both answers are valid.

#x = 2# and #x = -1#