One process is to first put each fraction over a common denominator (which for this problem would be #15#) by multiplying each fraction by the appropriate form of #1#:
#5/5((x - 8)/3) = 3/3 xx x/5#
#(5(x - 8))/15 = (3x)/15#
#((5 xx x) - (5 xx 8))/15 = (3x)/15#
#(5x - 40)/15 = (3x)/15#
#(5x)/15 - 40/15 = (3x)/15#
Next, add #color(red)(40/15)# and subtract #color(blue)((3x)/15)# from each side of the equation to isolate the #x# terms while keeping the equation balanced:
#-color(blue)((3x)/15) + (5x)/15 - 40/15 + color(red)(40/15) = -color(blue)((3x)/15) + (3x)/15 + color(red)(40/15)#
#(-color(blue)((3x)) + 5x)/15 - 0 = 0 + color(red)(40/15)#
#((-color(blue)(3) + 5)x)/15 = 40/15#
#(2x)/15 = 40/15#
Now, multiply each side of the equation by #color(red)(15)/color(blue)(2)# to solve for #x# while keeping the equation balanced:
#color(red)(15)/color(blue)(2) xx (2x)/15 = color(red)(15)/color(blue)(2) xx 40/15#
#cancel(color(red)(15))/cancel(color(blue)(2)) xx (color(blue)(cancel(color(black)(2)))x)/color(red)(cancel(color(black)(15))) = cancel(color(red)(15))/cancel(color(blue)(2)) xx (color(blue)(cancel(color(black)(40)))20)/color(red)(cancel(color(black)(15)))#
#x = 20#
