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

2 Answers
Nov 26, 2017

Answer:

#x=1/3" or "x=-5/3#

Explanation:

#"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#

Nov 26, 2017

Answer:

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

Explanation:

The piecewise definition of the absolute value function is:

#|f(x)| = {(f(x); f(x) >= 0),(-f(x); f(x) < 0):}#

In this case #f(x) = 2(1/3+1/2x)#

Substitute into 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):}#

Simplify the domain restrictions:

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

Separate the given equation into two equations with its respective domain restriction:

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

Multiply the second equation by -1:

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

Distribute the two in both 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 domain restrictions can be dropped, because neither equation violates them:

#x = 1/3# and #x = -5/3#