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

Question

Isaiah Anthony · Accepted Answer

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]}

Ethan Adkins · Answer

undefined To solve $x + 3 = -|3x - 1|$, you first isolate the absolute value expression on one side of the equation. Then, you solve for $x$ by considering both the positive and negative cases for the absolute value expression. After solving, you get $x = -2$ or $x = \frac{4}{3}$.