First, multiply each side of the equation by #color(red)(5)(color(blue)(3b + 2))# to eliminate the fraction while keeping the equation balanced:

#color(red)(5)(color(blue)(3b + 2)) xx 6/(3b + 2) = color(red)(5)(color(blue)(3b + 2)) xx 3/5#

#color(red)(5)cancel((color(blue)(3b + 2))) xx 6/color(blue)(cancel(color(black)((3b + 2)))) = cancel(color(red)(5))(color(blue)(3b + 2)) xx 3/color(red)(cancel(color(black)(5)))#

#5 xx 6 = (3b + 2) x 3#

#30 = (3b xx 3) + (2 xx 3)#

#30 = 9b + 6#

Next, subtract #color(red)(6)# from each side of the equation to isolate the #b# term while keeping the equation balanced:

#30 - color(red)(6) = 9b + 6 - color(red)(6)#

#24 = 9b + 0#

#24 = 9b#

Now, divide each side of the equation by #color(red)(9)# to solve for #b# while keeping the equation balanced:

#24/color(red)(9) = (9b)/color(red)(9)#

#(3 xx 8)/(3 xx 3) = (color(red)(cancel(color(black)(9)))b)/cancel(color(red)(9))#

#(cancel(3) xx 8)/(cancel(3) xx 3) = b#

#8/3 = b#

#b = 8/3#