First, add #color(red)(3)# and #color(blue)(4.3b)# to each side of the equation to isolate the #b# term while keeping the equation balanced:
#13.7b + color(blue)(4.3b) - 3 + color(red)(3) = 3 + color(red)(3) - 4.3b + color(blue)(4.3b)#
#(13.7 + color(blue)(4.3))b - 0 = 6 - 0#
#18b = 6#
Now, divide each side of the equation by #color(red)(18)# to solve for #b# while keeping the equation balanced:
#(18b)/color(red)(18) = 6/color(red)(18)#
#(color(red)(cancel(color(black)(18)))b)/cancel(color(red)(18)) = 6/color(red)(6 xx 3)#
#b = color(red)(cancel(color(black)(6)))/color(red)(color(black)(cancel(color(red)(6))) xx 3)#
#b = 1/3#