How do you solve #abs(3-x)-1=abs(4x+2)#?

1 Answer
Jul 17, 2015

Answer:

#color(red)(x = 0)#

Explanation:

We have two absolute conditions, so we must write four different equations without the absolute value symbols and solve for #x#.

Your original equation is

#|3-x|-1 = |4x+2|#

So these equations would be

(1): #(3-x) -1 = (4x+2)#
(2): #(3-x) - 1 = -(4x+2)#
(3): #-(3-x) - 1 = (4x+2)#
(4): #-(3-x) - 1 = -(4x+2)#

Solve Equation 1:

#(3-x) -1 = (4x+2)#
#3-x-1 = 4x+2#
#2-x = 4x+2#
#-x = 4x#
#0 = 5x#
#x = 0#

This is a possible solution, which we will check later.

Solve Equation 2.

#(3-x) -1 = -(4x+2)#
#3-x-1 = -4x-2#
#2-x = -4x-2#
#-x = -4x#
#0 = -3x#
#x = 0#

This is a possible solution, which we will check later.

Solve Equation 3.

#-(3-x) - 1 = (4x+2)#
#-3+x-1 = 4x+2#
#x-4 = 4x+2#
#3x = -6#
#x = -2#

This is a possible solution, which we will check later.

Solve Equation 4.

#-(3-x) - 1 = -(4x+2)#
#-3+x-1 = -4x-2#
#x-4 = -4x-2#
#5x = 2#
#x = 2/5#

This is a possible solution, which we will check later.

Check Solutions 1 and 2

If #x=0#

#|3-x|-1 = |4x+2|#
#|3-0|-1 = |4×0+2|#
#|3|-1 =|0+2|#
#3-1= |2|#
#2 = 2#

Check Solution 3

If #x = -2#

#|3-x|-1 = |4x+2|#
#|3-(-2)|-1 = |4(-2)+2|#
#|3+2| -1 = |-8 +2|#
#|5| -1 = |-6|#
#5-1 = 6#
#4 = 6#

This is an impossible result. #x = -2# is not a solution.

Check Solution 4

If #x = 2/5#

#|3-x|-1 = |4x+2|#
#|3-2/5|-1 = |4×2/5+2|#
#|15/5-2/5|-1 = |8/5+10/5|#
#|13/5|-1 = |18/5|#
#13/5 – 1 = 18/5#
#13/5- 5/5 = 18/5#
#8/5 = 18/5#

This is an impossible result. #x = 2/5# is not a solution.

The only solution is #x = 0#.