How do you solve #2-3abs(x-1)=-4abs(x-1)+7#?

(1) #x>=1# so #x-1# is non-negative
The equality turns into:
#2-3(x-1)=-4(x-1)+7->#
#2-3x+3=-4x+4+7->#
#-3x+4x=4+7-2-3->#
#x=6#
(and this is #>=1#) -- always check this!

(2) #x<1# so #x-1# is negative. The absolute bars will turn the sign around and the equality turns into:
#2-3(1-x)=-4(1-x)+7->#
#2-3+3x=-4+4x+7->#
#3x-4x=-4+7-2+3->#
#-x=4 -> x=-4#
(and this is #<1#)

#2-3abs(x-1)=-4abs(x-1)+7#

Observe that the expression inside the absolute value signs is the same in both absolute values. It is #x-1#. This approach will first find out what #abs(x-1)# is equal to, and then, find out what #x# must be.

Until you get some experience with working with expressions as if they were variables, it will help to actually do a substitution.
Our first goal is to find #abs(x-1)#. Since this is an absolute value, let's call it #a# for the time being:

Let #a = abs(x-1)#.

Substituting, we have:

#2-3a=-4a+7# Find #a#. (add #4a# and subtract #2# on both sides)

#-3a+4a=7-2# (simplify)

#a = 5#

Good. We've finished step 1. Now we still need to find #x#. We'll use #a=abs(x-1)# to write:

#abs(x-1) = 5#

The two numbers whose absolute value is #5# are #-5# and #5#,, so

#x-1=-5# #color(white)"xx"# or #color(white)"xx"# #x-1 = 5# #color(white)"xx"# so

#x=-4# #color(white)"xx"# or #color(white)"xx"# #x = 6#

Less details looks like this:

#2-3abs(x-1)=-4abs(x-1)+7#

#-3abs(x-1) +4abs(x-1) = 7-2#

#abs(x-1) =5#

#x-1 = -5# or #x-1=5#

#x=-4 " or " 6#

Solutions: #-4#, and #6#