First, expand the terms within the parenthesis on the left side of the equation by multiplying each term within the parenthesis by the term outside the parenthesis:
#color(red)(2/5)(x - 4) = 2x#
#(color(red)(2/5) xx x) - (color(red)(2/5) xx 4) = 2x#
#2/5x - 8/5 = 2x#
Next, subtract #color(red)(2/5x)# from each side of the equation to isolate the #x# term while keeping the equation balanced:
#-color(red)(2/5x) + 2/5x - 8/5 = 2x - color(red)(2/5x)#
#0 - 8/5 = (5/5 xx 2x) - color(red)(2/5x)#
#-8/5 = 10/5x - color(red)(2/5x)#
#-8/5 = (10/5 - color(red)(2/5))x#
#-8/5 = 8/5x#
Now, multiply each side of the equation by #color(red)(5/8)# to solve for #x# while keeping the equation balanced:
#color(red)(5/8) xx -8/5 = color(red)(5/8) xx 8/5x#
#-40/40 = 40/40x#
#-1 = 1x#
#-1 = x#
#x = -1#
