First, subtract #color(red)(w/3)# from each side of the equation to isolate the #w# term while keeping the equation balanced:
#-color(red)(w/3) + w/3 + 2 = -color(red)(w/3) + w/2#
#0 + 2 = (2/2 xx -color(red)(w/3)) + (3/3 xx w/2)#
#2 = (-2w)/6 + (3w)/6#
#2 = (-2w + 3w)/6#
#2 = ((-2 + 3)w)/6#
#2 = (1w)/6#
#2 = w/6#
Now, multiply each side of the equation by #color(red)(6)# to solve for #w# while keeping the equation balanced:
#color(red)(6) xx 2 = color(red)(6) xx w/6#
#12 = cancel(color(red)(6)) xx w/color(red)(cancel(color(black)(6)))#
#12 = w#
#w = 12#
