How do you solve the equation #abs(2(1/3+1/2x))=1#?

Question

Ethan Adkins · Accepted Answer

#x=1/3" or "x=-5/3# #"distribute the factor"# #rArr|(2(1/3+1/2x)|=|2/3+x|# #"the value inside the absolute value function can be"# #color(blue)"positive or negative"# #color(blue)"first solution"# #2/3+x=1rArrx=1-2/3=1/3# #color(blue)"second solution"# #-(2/3+x)=1# #rArr-2/3-x=1# #rArr-x=1+2/3=5/3rArrx=-5/3# #color(blue)"As a check"# #x=1/3"# #rArr|2/3+1/3|=|1|=1# #x=-5/3# #rArr|2/3-5/3|=|-1|=1#

Nathan Axel · Answer

Use the piecewise definition of the absolute value function to separate the equation into two equations and then solve each equation.

The absolute value function can be defined piecewise as follows: #|f(x)| = {(f(x); f(x) >= 0),(-f(x); f(x) < 0):}# In this case #f(x) = 2(1/3+1/2x)# Replace in the definition: #|2(1/3+1/2x)| = {(2(1/3+1/2x); 2(1/3+1/2x) >= 0),(-2(1/3+1/2x); 2(1/3+1/2x) < 0):}# Make the domain restrictions simpler: #|2(1/3+1/2x)| = {(2(1/3+1/2x); x >= -2/3),(-2(1/3+1/2x); x < -2/3):}# Divide the provided equation into two parts, each with a corresponding domain restriction: #2(1/3+1/2x) = 1; x >=-2/3# and #-2(1/3+1/2x) = 1; x < -2/3# Multiply by -1 to get the second equation: #2(1/3+1/2x) = 1; x >=-2/3# and #2(1/3+1/2x) = -1; x < -2/3# Divide the two between the two equations: #2/3+x = 1; x >=-2/3# and #2/3+x = -1; x < -2/3# Subtract #2/3# from both sides of both equations: #x = 1/3; x >=-2/3# and #x = -5/3; x < -2/3# The restrictions on the domain can be removed, as neither equation goes against them: #x = 1/3# and #x = -5/3#

Isaiah Anthony · Answer

undefined To solve the equation $ |2\left(\frac{1}{3} + \frac{1}{2}x\right)| = 1 $, first, isolate the absolute value expression. Then, split the equation into two cases: when the expression inside the absolute value is positive and when it is negative. Solve each case separately to find the values of $ x $.