First, multiply both sides of the equation by #color(red)(10)# to eliminate the fractions while keeping the equation balanced. We use #10# because it is the LCD (lowest common denominator) for all of the fractions:
#color(red)(10) xx (2x - 3)/10 = color(red)(10)(1/2x + 3/5)#
#cancel(color(red)(10)) xx (2x - 3)/color(red)(cancel(color(black)(10))) = (color(red)(10) xx 1/2x) + (color(red)(10) xx 3/5)#
#2x - 3 = (cancel(color(red)(10))5 xx 1/color(red)(cancel(color(black)(2)))x) + (cancel(color(red)(10))2 xx 3/color(red)(cancel(color(black)(5))))#
#2x - 3 = 5x + 6#
Next, subtract #color(red)(2x)# and #color(blue)(6)# from each side of the equation to isolate the #x# term while keeping the equation balanced:
#2x - 3 - color(red)(2x) - color(blue)(6) = 5x + 6 - color(red)(2x) - color(blue)(6)#
#2x - color(red)(2x) - 3 - color(blue)(6) = 5x - color(red)(2x) + 6 - color(blue)(6)#
#0 - 9 = (5 - 2)x + 0#
#-9 = 3x#
Now, divide each side of the equation by #color(red)(3)# to solve for #x# while keeping the equation balanced:
#-9/color(red)(3) = (3x)/color(red)(3)#
#-3 = (color(red)(cancel(color(black)(3)))x)/cancel(color(red)(3))#
#-3 = x#
#x = -3#
