First, we'll re-arrange the equation such that all constants are on one side, and all of the #x#-related terms are on the other. We'll do this by adding and subtracting terms from both sides so terms will cancel out on one side, but modify the other:
First, we'll move the #4/5x# term:
#cancel(-4/5x)+6color(red)(cancel(+4/5x))=x-3color(red)(+4/5x)#
#6=x(1+color(red)(4/5))-3#
#6=9/5x-3#
Next, we'll move the -3 constant to the left-hand side:
#6color(blue)(+3)=9/5xcancel(-3)color(blue)(cancel(+3))#
#9=9/5x#
Now, all that's left is to divide both sides by #x#'s coefficient. Because we're dividing a fraction, we can write it as both sides being multiplied by its inverse:
#cancel(9)color(red)(xx5/cancel(9))=cancel(9/5)x xxcolor(red)(cancel(5/9))#
#color(green)(x=5)#
