First, subtract #color(red)(4)# and #color(blue)(1/4d)# from each side of the equation to isolate the #d# term while keeping the equation balanced:
#3/4d + 4 - color(red)(4) - color(blue)(1/4d) = 1/4d + 18 - color(red)(4) - color(blue)(1/4d)#
#3/4d - color(blue)(1/4d) + 4 - color(red)(4) = 1/4d - color(blue)(1/4d) + 18 - color(red)(4)#
#(3 - 1)/4d + 0 = 0 + 14#
#2/4d = 14#
#1/2d = 14#
Now, multiply each side of the equation by #color(red)(2)# to solve for #d# while keeping the equation balanced:
#color(red)(2) xx 1/2d = color(red)(2) xx 14#
#cancel(color(red)(2)) xx 1/color(red)(cancel(color(black)(2)))d = 28#
#1d = 28#
#d = 28#
