First, expand the terms within parenthesis on both sides of the equation:

#(3 xx 3u) + (3 xx 2) + 5 = (2 xx 2u) - (2 xx 2)#

#9u + 6 + 5 = 4u - 4#

#9u + 11 = 4u - 4#

Next, subtract #color(red)(4u)# and #color(blue)(11)# from each side of the equation to isolate the #u# term while keeping the equation balanced:

#9u + 11 - color(red)(4u) - color(blue)(11) = 4u - 4 - color(red)(4u) - color(blue)(11)#

#9u - color(red)(4u) + 11 - color(blue)(11) = 4u - color(red)(4u) - 4 - color(blue)(11)#

#(9 - 4)u + 0 = 0 - 15#

#5u = -15#

Now, divide each side of the equation by #color(red)(5)# to solve for #u# while keeping the equation balanced:

#(5u)/color(red)(5) = (-15)/color(red)(5)#

#(color(red)(cancel(color(black)(5)))u)/cancel(color(red)(5)) = -3#

#u = -3#