First, multiply both sides of the equation by #color(red)(15)# to eliminate the fractions to make the problem easier to work with and to keep the equation balanced:

#color(red)(15)(3/5x - 2) = color(red)(15) xx 1/3#

#(color(red)(15) xx 3/5x) - (color(red)(15) xx 2) = cancel(color(red)(15))5 xx 1/color(red)(cancel(color(black)(3)))#

#(cancel(color(red)(15))3 xx 3/color(red)(cancel(color(black)(5)))x) - 30 = 5#

#9x - 30 = 5#

Next, add #color(red)(30)# to each side of the equation to isolate the #x# term while keeping the equation balanced:

#9x - 30 + color(red)(30) = 5 + color(red)(30)#

#9x - 0 = 35#

#9x = 35#

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

#(9x)/color(red)(9) = 35/color(red)(9)#

#(color(red)(cancel(color(black)(9)))x)/cancel(color(red)(9)) = 35/9#

#x = 35/9#