How do you solve # |x – 4| > |3x – 1|#?

When there is absolute value, we have to take two options the first where the first is positive the next is negative.
#|a+b|#
#=a+b#
or #=-(a+b)=-a-b#
#|x–4|>|3x–1|#

1. #=x-4>|3x-1|#
again doing the same with the left hand side:
#x-4>|3x-1|#

1a. #=x-4>3x-1#

Subtracting x from both sides of the equation to get the unknown's terms to one side of the equation:
#cancelxcancel(-x)-4> 3x-1-x#
#-4> 2x-1#

Adding 1 to both sides of the equation to isolate the unknown's term on the left hand side:
#-4+1> 2xcancel(-1)cancel(+1)#
#-3> 2x#

Dividing both sides of the equation by 2 to isolate x:
#-3/2> (cancel2 x)/(cancel2)#

1a. #color(red)(x<-3/2#

and

1b. #x-4 > -(3x-1)#
#=x-4> -3x+1#
Doing same as before to isolate x:
#cancelx-4cancel(-x)> -3x+1-x#
#-4> -4x#
#(cancel(-4))/cancel(-4)> (cancel(-4)x)/cancel(-4)#
#1>x#
1b. #color(red)(x<1#

and

1. #=-(x-4)>|3x-1|#
   #=-x+4>|3x-1|#

2a.#=-x+4>3x-1#

simplifying: (adding 1 and subtracting x from both sides of the equation and then dividing by 4):
#cancel(-x)cancel(+x)+4+1>3xcancel(-1)+xcancel(+1)#
#4+1>3x+x#
#5>4x#

#5/4>(cancel4 x)/(cancel4)#
#5/4>x#
2a.#color(red)(x<5/4#

and

2b.#=-x+4> -1(3x-1)#
#=-x+4> -3x+1#

Simplifying by adding 3x to, subtracting 4 from both sides and then dividing by 2:
#-xcancel(+4)+3xcancel(-4)> cancel(-3x)+1cancel(+3x)-4#
#-x+3x>1-4#
#2x>-3#
#(cancel2x)/cancel 2> -3/2#
2b. #color(red)(x> -3/2)#

Simplifying 1a , 2a ,and 2b :
#x< -3/2#
#x<1#
#x<5/4rarrx<1.25#
#rarr x<-3/2<1<5/4#
#rarr color(blue)(x<5/4#

and

From 2a:
#color(blue)(x> -3/2#

since x is smaller than #5/4# and bigger than #-3/2# then x defined for any number.