First, add #color(red)(4)# and subtract #color(blue)(a)# to each side of the equation to isolate the #a# term while keeping the equation balanced:
#4a - 4 + color(red)(4) - color(blue)(a) = 8 + a + color(red)(4) - color(blue)(a)#
#4a - color(blue)(a) - 4 + color(red)(4) = 8 + color(red)(4) + a - color(blue)(a)#
#4a - color(blue)(1a) - 0 = 12 + 0#
#(4 - color(blue)(1))a = 12#
#3a = 12#
Now, divide each side of the equation by #color(red)(3)# to solve for #a# while keeping the equation balanced:
#(3a)/color(red)(3) = 12/color(red)(3)#
#(color(red)(cancel(color(black)(3)))a)/cancel(color(red)(3)) = 4#
#a = 4#