First, subtract #color(red)((2x)/3)# and add #color(blue)(4)# to each side of the inequality to isolate the #x# term while keeping the inequality balanced:
#-color(red)((2x)/3) + (2x)/3 - 2 + color(blue)(4) < color(red)((2x)/3) + x - 4 + color(blue)(4)#
#0 + 2 < -color(red)((2x)/3) + 1x - 0#
#2 < -color(red)((2x)/3) + 3/3x#
#2 < (-color(red)(2/3) + 3/3)x#
#2 < 1/3x#
Now, multiply each side of the inequality by #color(red)(3)# to solve for #x# while keeping the inequality balanced:
#color(red)(3) xx 2 < color(red)(3) xx 1/3x#
#6 < cancel(color(red)(3)) xx 1/color(red)(cancel(color(black)(3)))x#
#6 < x#
To state the solution in terms of #x# we can reverse or "flip" the entire inequality:
#x > 6#