First, subtract #color(red)(5)# and #color(blue)(4z)# from each side of the equation to isolate the #z# terms while keeping the equation balanced:
#5 - 8z - color(red)(5) - color(blue)(4z) = 25 + 4z - color(red)(5) - color(blue)(4z)#
#5 - color(red)(5) - 8z - color(blue)(4z) = 25 - color(red)(5) + 4z - color(blue)(4z)#
#0 - 12z = 20 + 0#
#-12z = 20#
Now, divide each side of the equation by #color(red)(-12)# to solve for #z# while keeping the equation balanced:
#(-12z)/color(red)(-12) = 20/color(red)(-12)#
#(color(red)(cancel(color(black)(-12)))z)/cancel(color(red)(-12)) = (4 xx 5)/color(red)(4 xx -3)#
#z = (color(red)(cancel(color(black)(4))) xx 5)/color(red)(cancel(4) xx -3)#
#z = 5/-3#
#z = -5/3#