First, multiply each side of the inequality by #color(red)(9)# to eliminate the fractions while keeping the inequality balanced:
#color(red)(9) xx (x + 3)/9 > color(red)(9) xx (x - 1)/3#
#cancel(color(red)(9)) xx (x + 3)/color(red)(cancel(color(black)(9))) > cancel(color(red)(9))3 xx (x - 1)/color(red)(cancel(color(black)(3)))#
#x + 3 > 3(x - 1)#
#x + 3 > (3 xx x) - (3 xx 1)#
#x + 3 > 3x - 3#
Next, subtract #color(red)(x)# and add #color(blue)(3)# to each side of the inequality to isolate the #x# term while keeping the inequality balanced:
#-color(red)(x) + x + 3 + color(blue)(3) > -color(red)(x) + 3x - 3 + color(blue)(3)#
#0 + 6 > -color(red)(1x) + 3x - 0#
#6 > (-color(red)(1) + 3)x#
#6 > 2x#
Now, divide each side of the inequality by #color(red)(2)# to solve for #x# while keeping the inequality balanced:
#6/color(red)(2) > (2x)/color(red)(2)#
#3 > (color(red)(cancel(color(black)(2)))x)/cancel(color(red)(2))#
#3 > x#
We can reverse or "flip" the entire inequality to state the solution in terms of #x#:
#x < 3#
