How do you solve #abs(2x-4)>10#?

Here are two methods you might use to solve this:

Method 1: Start by squaring both sides
#abs(2x-4) > 10#

#color(white)("XXX")rarr (2x-4)^2 > 10^2#

#color(white)("XXX")4x^2-16x+16 > 100#

#color(white)("XXX")x^2-4x-21 > 0#

#color(white)("XXX")(x-7)(x+3) > 0#

This will be true it both #(x-7)# and #(x+3)# have the same sign.
That is if
#color(white)("XXX")(x-7) < 0 rarr x < 7#
#color(white)("XXX")#and
#color(white)("XXX")(x+3) < 0 rarr x < -3#
#color(white)("XXXXXXXXX")#for both to be true: #x < -3#
or
#color(white)("XXX")(x-7) > 0 rarr x > 7#
#color(white)("XXX")#and
#color(white)("XXX")(x+3) > 0 rarr x > -3#
#color(white)("XXXXXXXXX")#for both to be true: #x > 7#

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Method 2: Separate the negative and positive versions of #(2x-4)# immediately
#abs(2x-4) > 0#

#{: ("if " (2x -4) < 0,color(white)("X")orcolor(white)("X"),"if " (2x-4) > 0), (abs(2x-4) > 10 rarr 4-2x > 10,,abs(2x-4) > 10 rarr 2x-4 > 10), (color(white)("XXXXXXXX")rarr -2x > 6,,color(white)("XXXXXXXX")rarr x > 7), (color(white)("XXXXXXXX")rarr x < -3,,) :}#

#|2x-4|>10#

If #|2x-4|# is negative,

#-(2x-4)>10#
#color(white)(,)-2x+4>10#
#color(white)(xxx.)-2x>6#
#color(white)(xxxxx.)2x<-6#
#color(white)(xxxxx..)x<-3#

If #|2x-4|# is positive,

#2x-4>10#
#color(white)(xxx)2x>14#
#color(white)(xxx.)x>7#

Hence #x<-3# or #x>7#.