First, add #color(red)(5)# and subtract #color(blue)(x/3)# from both sides of the equation to isolate the #x# term while keeping the equation balanced:
#(2x)/3 - 5 + color(red)(5) - color(blue)(x/3) = x/3 + 2 + color(red)(5) - color(blue)(x/3)#
#(2x)/3 - color(blue)(x/3) - 5 + color(red)(5) = x/3 - color(blue)(x/3) + 2 + color(red)(5)#
#(2x - x)/3 - 0 = 0 + 7#
#(2x - 1x)/3 = 7#
#(1x)/3 = 7#
#x/3 = 7#
Now, multiply each side of the equation by #color(red)(3)# to solve for #x# while keeping the equation balanced:
#color(red)(3) xx x/3 = color(red)(3) xx 7#
#cancel(color(red)(3)) xx x/color(red)(cancel(color(black)(3))) = 21#
#x = 21#