First, expand the terms in parenthesis on each side of the equation by multiplying each term within the parenthesis by the term outside the parenthesis:

#color(red)(9)(4b - 1) = color(blue)(2)(9b + 3)#

#(color(red)(9) * 4b) - (color(red)(9) * 1) = (color(blue)(2) * 9b) + (color(blue)(2) * 3)#

#36b - 9 = 18b + 6#

Next, add #color(red)(9)# and subtract #color(blue)(18b)# from each side of the equation to isolate the #b# term while keeping the equation balanced:

#36b - 9 + color(red)(9) - color(blue)(18b) = 18b + 6 + color(red)(9) - color(blue)(18b)#

#36b - color(blue)(18b) - 9 + color(red)(9) = 18b - color(blue)(18b) + 6 + color(red)(9)#

#(36 - color(blue)(18))b - 0 = 0 + 15#

#18b = 15#

Now, divide each side of the equation by #color(red)(18)# to solve for #b# while keeping the equation balanced:

#(18b)/color(red)(18) = 15/color(red)(18)#

#(color(red)(cancel(color(black)(18)))b)/cancel(color(red)(18)) = (3 xx 5)/color(red)(3 xx 6)#

#b = (color(red)(cancel(color(black)(3))) xx 5)/color(red)(color(black)(cancel(color(red)(3))) xx 6)#

#b = 5/6#