First, expand the terms within parenthesis:
#(color(red)(-2) xx x) - (color(red)(-2) xx 4) <= 3x#
#-2x + 8 <= 3x#
Next, add #color(red)(2x)# to each side of the inequality to isolate the #x# term while keeping the inequality balanced:
#color(red)(2x) - 2x + 8 <= color(red)(2x) + 3x#
#0 + 8 <= (2 + 3)x#
#8 <= 5x#
Now, divide each side of the inequality by #color(red)(5)# to solve for #x# while keeping the inequality balanced:
#8/color(red)(5) <= (5x)/color(red)(5)#
#8/5 <= (color(red)(cancel(color(black)(5)))x)/cancel(color(red)(5))#
#8/5 <= x#
To put the solution in terms of #x# we can reverse or "flip" the inequality:
#x >= 8/5#